Past Exam Questions

Answer: (i) \(\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}=\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\).

(ii) \(S_N=\dfrac{1}{\mathrm e}-\dfrac{1}{(N+1)\mathrm e^{N+1}}\), so \(S=\dfrac{1}{\mathrm e}\).

(iii) The least value is \(N=6\).

For the identity, put the right-hand side over the common denominator \(n(n+1)\mathrm e^{n+1}\):

\(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}=\dfrac{(n+1)\mathrm e-n}{n(n+1)\mathrm e^{n+1}}=\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}.\)

Therefore

\(S_N=\sum_{n=1}^{N}\left(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\right).\)

This is a telescoping sum:

\(S_N=\left(\dfrac1{\mathrm e}-\dfrac1{2\mathrm e^2}\right)+\left(\dfrac1{2\mathrm e^2}-\dfrac1{3\mathrm e^3}\right)+\cdots+\left(\dfrac1{N\mathrm e^N}-\dfrac1{(N+1)\mathrm e^{N+1}}\right).\)

All intermediate terms cancel, so

\(S_N=\dfrac1{\mathrm e}-\dfrac1{(N+1)\mathrm e^{N+1}}.\)

Hence

\(S=\lim_{N\to\infty}S_N=\dfrac1{\mathrm e}.\)

Then

\(S-S_N=\dfrac1{(N+1)\mathrm e^{N+1}}\),

so

\((N+1)(S-S_N)=\dfrac1{\mathrm e^{N+1}}.\)

We need \(\mathrm e^{-(N+1)}\lt 10^{-3}\). Since \(\ln 1000\approx 6.91\), we need \(N+1>6.91\), so the least integer value is

\(N=6\).

Answer: (i) \(S_N = N(3N^2+12N+13)\).

(ii) \(T_N = \frac{1}{12}-\frac{1}{3(3N+4)}\).

(iii) \(\frac{S_N}{T_N} = 4(3N+4)(3N^2+12N+13)\), so it is an integer.

(iv) \(\lim_{N\to\infty}\frac{S_N}{N^3T_N}=36\).

(i) Expand the term in the sum:

\((3r+1)(3r+4)=9r^2+15r+4\).

So

\(S_N=\sum_{r=1}^N (9r^2+15r+4)=9\sum_{r=1}^N r^2+15\sum_{r=1}^N r+4\sum_{r=1}^N 1\).

Using the standard formulae

\(\sum_{r=1}^N r=\frac{1}{2}N(N+1)\), \(\sum_{r=1}^N r^2=\frac{1}{6}N(N+1)(2N+1)\), and \(\sum_{r=1}^N 1=N\),

we get

\(S_N=9\left(\frac{1}{6}N(N+1)(2N+1)\right)+15\left(\frac{1}{2}N(N+1)\right)+4N\).

Now simplify:

\(S_N=N\left(\frac{9}{6}(N+1)(2N+1)+\frac{15}{2}(N+1)+4\right)\).

Expanding inside the bracket gives

\(S_N=N(3N^2+12N+13)\).

(ii) Use partial fractions:

\(\frac{1}{(3r+1)(3r+4)}=\frac{A}{3r+1}+\frac{B}{3r+4}\).

Solving gives \(A=\frac13\) and \(B=-\frac13\), so

\(\frac{1}{(3r+1)(3r+4)}=\frac13\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right)\).

Hence

\(T_N=\frac13\sum_{r=1}^N\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right)\).

Writing out the first few terms,

\(T_N=\frac13\left(\left(\frac14-\frac17\right)+\left(\frac17-\frac{1}{10}\right)+\cdots+\left(\frac{1}{3N+1}-\frac{1}{3N+4}\right)\right)\).

Everything cancels except the first positive term and the last negative term, so

\(T_N=\frac13\left(\frac14-\frac{1}{3N+4}\right)=\frac{1}{12}-\frac{1}{3(3N+4)}\).

(iii) From (ii),

\(T_N=\frac13\left(\frac14-\frac{1}{3N+4}\right)=\frac{1}{3}\cdot\frac{3N}{4(3N+4)}=\frac{N}{4(3N+4)}\).

Therefore

\(\frac{S_N}{T_N}=\frac{N(3N^2+12N+13)}{N/(4(3N+4))}=4(3N+4)(3N^2+12N+13)\).

This is a product of integers, so \(\frac{S_N}{T_N}\) is an integer.

(iv) Using the expression above,

\(\frac{S_N}{N^3T_N}=\frac{4(3N+4)(3N^2+12N+13)}{N^3}\).

As \(N\to\infty\), the leading terms dominate:

\(4(3N+4)(3N^2+12N+13)/N^3 \to 4\cdot 3\cdot 3=36\).

So the limit is \(36\).

Answer: (i) \(\sum_{r=n+1}^{2n} u_r = 14n^3 + 9n^2\).

(ii) \(u_r = 6r^2 - 1\).

Let \(S_n = \sum_{r=1}^{n} u_r\). We are given that \(S_n = n^2(2n+3)\).

(i) The sum from \(r=n+1\) to \(r=2n\) is

\(\sum_{r=n+1}^{2n} u_r = S_{2n} - S_n\).

Now

\(S_{2n} = (2n)^2(2(2n)+3) = 4n^2(4n+3) = 16n^3 + 12n^2\).

So

\(S_{2n} - S_n = (16n^3 + 12n^2) - (2n^3 + 3n^2) = 14n^3 + 9n^2\).

Hence \(\sum_{r=n+1}^{2n} u_r = 14n^3 + 9n^2\).

(ii) Since \(u_r\) is the difference between consecutive partial sums,

\(u_r = S_r - S_{r-1}\).

Now

\(S_r = r^2(2r+3) = 2r^3 + 3r^2\),

and

\(S_{r-1} = (r-1)^2\bigl(2(r-1)+3\bigr) = (r-1)^2(2r+1)\).

So

\(u_r = r^2(2r+3) - (r-1)^2(2r+1)\).

Expanding,

\(r^2(2r+3) = 2r^3 + 3r^2\),

\((r-1)^2(2r+1) = (r^2 - 2r + 1)(2r+1) = 2r^3 - 3r^2 + 1\).

Therefore

\(u_r = (2r^3 + 3r^2) - (2r^3 - 3r^2 + 1) = 6r^2 - 1\).

Answer: (i) Expanding the right-hand side gives

\(\frac{1}{2}\left\{\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right\}\)

\(=\frac{1}{2}\left\{\frac{(2r+1)(2r+3)r-(2r-1)(2r+1)(r+2)}{r(r+1)(r+2)}\right\}.\)

Now

\((2r+1)(2r+3)r= r(4r^2+8r+3)=4r^3+8r^2+3r,\)

and

\((2r-1)(2r+1)(r+2)=(4r^2-1)(r+2)=4r^3+8r^2-r-2.\)

So the numerator becomes

\(4r^3+8r^2+3r-(4r^3+8r^2-r-2)=4r+2.\)

Hence the right-hand side is

\(\frac{1}{2}\cdot\frac{4r+2}{r(r+1)(r+2)}=\frac{2r+1}{r(r+1)(r+2)},\)

as required.

(ii) Therefore

\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}\)

\(=\frac12\sum_{r=1}^{n}\left(\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right).\)

This is telescoping:

\(\frac12\Bigg[\left(\frac{3\cdot 5}{2\cdot 3}-\frac{1\cdot 3}{1\cdot 2}\right)+\left(\frac{5\cdot 7}{3\cdot 4}-\frac{3\cdot 5}{2\cdot 3}\right)+\cdots+\left(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{(2n-1)(2n+1)}{n(n+1)}\right)\Bigg].\)

All intermediate terms cancel, leaving

\(\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{1\cdot 3}{1\cdot 2}\right\}.\)

Since \(\frac{1\cdot 3}{1\cdot 2}=\frac32\),

\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac32\right\}.\)

(iii) As \(n\to\infty\),

\(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}\to 4.\)

So

\(\sum_{r=1}^{\infty}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left(4-\frac32\right)=\frac54.\)

(i) Start with the right-hand side and simplify the numerator:

\(\frac{1}{2}\left\{\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right\}\)

\(=\frac{1}{2}\left\{\frac{r(2r+1)(2r+3)-(r+2)(2r-1)(2r+1)}{r(r+1)(r+2)}\right\}.\)

Now expand:

\(r(2r+1)(2r+3)=r(4r^2+8r+3)=4r^3+8r^2+3r,\)

\((r+2)(2r-1)(2r+1)=(r+2)(4r^2-1)=4r^3+8r^2-r-2.\)

Hence the numerator is

\(4r^3+8r^2+3r-(4r^3+8r^2-r-2)=4r+2=2(2r+1).\)

Therefore

\(\frac{1}{2}\left\{\frac{2(2r+1)}{r(r+1)(r+2)}\right\}=\frac{2r+1}{r(r+1)(r+2)}.\)

This verifies the identity.

(ii) Using part (i),

\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\sum_{r=1}^{n}\left(\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right).\)

Write out the first few terms:

\(\frac12\Bigg[\left(\frac{3\cdot 5}{2\cdot 3}-\frac{1\cdot 3}{1\cdot 2}\right)+\left(\frac{5\cdot 7}{3\cdot 4}-\frac{3\cdot 5}{2\cdot 3}\right)+\cdots+\left(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{(2n-1)(2n+1)}{n(n+1)}\right)\Bigg].\)

Everything in the middle cancels, leaving

\(\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{1\cdot 3}{1\cdot 2}\right\}.\)

So

\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac32\right\}.\)

(iii) Now let \(n\to\infty\). Since

\(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}\to \frac{4n^2}{n^2}=4,\)

the infinite sum is

\(\frac12\left(4-\frac32\right)=\frac54.\)

Answer: \((r+1)^4-r^4=4r^3+6r^2+4r+1\).

Summing from \(r=1\) to \(r=n\) gives

\((n+1)^4-1=4\sum_{r=1}^{n}r^3+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n.\)

Using \(\sum r=\frac{n(n+1)}{2}\) and \(\sum r^2=\frac{n(n+1)(2n+1)}{6}\), this simplifies to

\(4\sum_{r=1}^{n}r^3=n^2(n+1)^2.\)

Therefore

\(\sum_{r=1}^{n}r^3=\frac14n^2(n+1)^2.\)

Expand the fourth power first:

\((r+1)^4=r^4+4r^3+6r^2+4r+1\), so

\((r+1)^4-r^4=4r^3+6r^2+4r+1\).

Now sum both sides from \(r=1\) to \(r=n\):

\(\sum_{r=1}^{n}\big((r+1)^4-r^4\big)=\sum_{r=1}^{n}(4r^3+6r^2+4r+1)\).

The left side telescopes:

\((2^4-1^4)+(3^4-2^4)+\cdots+((n+1)^4-n^4)=(n+1)^4-1^4\).

Therefore

\((n+1)^4-1=4\sum_{r=1}^{n}r^3+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n\).

Use the standard formulas

\(\sum_{r=1}^{n}r=\frac{n(n+1)}{2}\), \(\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}\).

Substitute these in:

\((n+1)^4-1=4\sum_{r=1}^{n}r^3+n(n+1)(2n+1)+2n(n+1)+n\).

Now simplify the right-hand extra terms:

\(n(n+1)(2n+1)=2n^3+3n^2+n\),

\(2n(n+1)=2n^2+2n\).

So

\((n+1)^4-1=4\sum_{r=1}^{n}r^3+2n^3+5n^2+4n\).

Also

\((n+1)^4-1=n^4+4n^3+6n^2+4n\).

Hence

\(n^4+4n^3+6n^2+4n=4\sum_{r=1}^{n}r^3+2n^3+5n^2+4n\).

Rearranging gives

\(4\sum_{r=1}^{n}r^3=n^4+2n^3+n^2=n^2(n^2+2n+1)=n^2(n+1)^2\).

Therefore

\(\sum_{r=1}^{n}r^3=\frac14 n^2(n+1)^2\).

Answer: Both sums have the same numerical value, namely \(2015\).

Using the formulae \(\sum_{r=1}^{n}r=\frac{n(n+1)}2\) and \(\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6\),

\(\sum_{r=1}^{13}(3r^2-5r+1)=3\sum_{r=1}^{13}r^2-5\sum_{r=1}^{13}r+13\).

So

\(\sum_{r=1}^{13}(3r^2-5r+1)=3\cdot\frac{13\cdot14\cdot27}{6}-5\cdot\frac{13\cdot14}{2}+13=2015\).

Also, since the term for \(r=0\) is \(-1\),

\(\sum_{r=0}^{9}(r^3-1)=\sum_{r=1}^{9}r^3-10\).

Using \(\sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}2\right)^2\),

\(\sum_{r=0}^{9}(r^3-1)=\left(\frac{9\cdot10}{2}\right)^2-10=45^2-10=2015\).

Therefore the two sums have the same numerical value.

Answer: \(\displaystyle \sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}\).

Therefore \(\displaystyle \sum_{r=1}^{\infty}\frac{1}{(2r)^2-1}=\frac12\).

First factor the denominator:

\((2r)^2-1=(2r-1)(2r+1).\)

Now use partial fractions:

\(\frac{1}{(2r-1)(2r+1)}=\frac12\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right).\)

Hence

\(\sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right).\)

Writing out the terms gives

\(\frac12\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right].\)

All the middle terms cancel, so

\(\sum_{r=1}^{n}\frac{1}{(2r)^2-1}=\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}.\)

Letting \(n\to\infty\),

\(\sum_{r=1}^{\infty}\frac{1}{(2r)^2-1}=\frac12.\)

Answer: \(\dfrac{1}{3}\left(\dfrac{1}{3r+1}-\dfrac{1}{3r+4}\right)=\dfrac{1}{(3r+1)(3r+4)}\).

\(S_N=\dfrac{1}{3}\left(\dfrac14-\dfrac{1}{3N+4}\right)\), so \(S=\dfrac{1}{12}\).

Therefore \(S-S_N=\dfrac{1}{3(3N+4)}\).

We need \(\dfrac{1}{3(3N+4)}\lt\dfrac{1}{10000}\), giving \(N\gt1109\dfrac{7}{9}\). Hence the least integer value is \(N=1110\).

Start with the right hand side:

\(\frac{1}{3}\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right)=\frac{1}{3}\left(\frac{(3r+4)-(3r+1)}{(3r+1)(3r+4)}\right)=\frac{1}{(3r+1)(3r+4)}.\)

Therefore

\(S_N=\sum_{r=1}^{N}\frac{1}{(3r+1)(3r+4)}=\frac{1}{3}\sum_{r=1}^{N}\left(\frac{1}{3r+1}-\frac{1}{3r+4}\right).\)

Writing out the terms gives a telescoping sum:

\(S_N=\frac{1}{3}\left[\left(\frac14-\frac17\right)+\left(\frac17-\frac1{10}\right)+\cdots+\left(\frac{1}{3N+1}-\frac{1}{3N+4}\right)\right].\)

All the middle terms cancel, so

\(S_N=\frac{1}{3}\left(\frac14-\frac{1}{3N+4}\right).\)

Taking the limit as \(N\to\infty\),

\(S=\frac{1}{3}\cdot\frac14=\frac{1}{12}.\)

Hence

\(S-S_N=\frac{1}{12}-\frac{1}{3}\left(\frac14-\frac{1}{3N+4}\right)=\frac{1}{3(3N+4)}.\)

We require

\(\frac{1}{3(3N+4)}\lt\frac{1}{10000}.\)

This gives

\(3(3N+4)\gt10000,\)

so

\(9N+12\gt10000,\qquad N\gt\frac{9988}{9}=1109\frac{7}{9}.\)

Therefore the least integer value of \(N\) is \(1110\).

Answer: The partial fraction decomposition is \(\frac{4}{r(r+1)(r+2)}=\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}\).

Hence

\(\sum_{r=1}^{n} \frac{4}{r(r+1)(r+2)}=1-\frac{2}{n+1}+\frac{2}{n+2}\).

So

\(\sum_{r=1}^{\infty} \frac{4}{r(r+1)(r+2)}=1\).

Let

\(\frac{4}{r(r+1)(r+2)}=\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}\).

Multiplying through by \(r(r+1)(r+2)\) gives

\(4=A(r+1)(r+2)+Br(r+2)+Cr(r+1).\)

Expanding:

\(4=A(r^2+3r+2)+B(r^2+2r)+C(r^2+r).\)

Collecting coefficients of powers of \(r\):

\(4=(A+B+C)r^2+(3A+2B+C)r+2A.\)

So

• \(A+B+C=0\)
• \(3A+2B+C=0\)
• \(2A=4\)

Hence \(A=2\). Then \(2+B+C=0\), so \(B+C=-2\), and \(6+2B+C=0\), so \(2B+C=-6\). Subtracting gives \(B=-4\), and therefore \(C=2\).

Thus

\(\frac{4}{r(r+1)(r+2)}=\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}.\)

Now sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n}\left(\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}\right).\)

Writing out the first few terms shows the cancellation:

\((2-2+\tfrac{2}{3})+(1-\tfrac{4}{3}+\tfrac{1}{2})+\cdots+(\tfrac{2}{n-1}-\tfrac{4}{n}+\tfrac{2}{n+1})+(\tfrac{2}{n}-\tfrac{4}{n+1}+\tfrac{2}{n+2})\).

All the interior terms cancel, leaving

\(\sum_{r=1}^{n} \frac{4}{r(r+1)(r+2)}=1-\frac{2}{n+1}+\frac{2}{n+2}.\)

Finally, as \(n\to\infty\), both fractions tend to \(0\), so

\(\sum_{r=1}^{\infty} \frac{4}{r(r+1)(r+2)}=1.\)

Answer: \(\displaystyle \sum_{r=1}^{n}(4r-3)(4r+1)=\frac{n}{3}(16n^{2}+12n-13)\).

First expand the bracket:

\((4r-3)(4r+1)=16r^{2}+4r-12r-3=16r^{2}-8r-3\).

So

\(\displaystyle \sum_{r=1}^{n}(4r-3)(4r+1)=\sum_{r=1}^{n}(16r^{2}-8r-3)\).

Split the sum into separate parts:

\(\displaystyle =16\sum_{r=1}^{n}r^{2}-8\sum_{r=1}^{n}r-3\sum_{r=1}^{n}1\).

Using the standard formulae

\(\displaystyle \sum_{r=1}^{n}r^{2}=\frac{n(n+1)(2n+1)}{6}\), \(\displaystyle \sum_{r=1}^{n}r=\frac{n(n+1)}{2}\), and \(\displaystyle \sum_{r=1}^{n}1=n\),

we get

\(\displaystyle =16\cdot \frac{n(n+1)(2n+1)}{6}-8\cdot \frac{n(n+1)}{2}-3n\).

Simplify:

\(\displaystyle =\frac{8}{3}n(n+1)(2n+1)-4n(n+1)-3n\).

Expanding and collecting terms gives

\(\displaystyle =\frac{n}{3}(16n^{2}+12n-13)\).

Answer: \(\sum_{k=13}^{n} u_k = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\).

Hence \(\sum_{k=13}^{\infty} u_k = \frac{1}{5}\).

Write out the first few terms:

\(u_{13}=\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}},\quad u_{14}=\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}},\quad u_{15}=\frac{1}{\sqrt{29}}-\frac{1}{\sqrt{31}}.\)

In general, the final term is

\(u_n=\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}.\)

So

\(\sum_{k=13}^{n} u_k=\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\cdots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right).\)

This is a telescoping sum, so all the middle terms cancel:

\(\sum_{k=13}^{n} u_k=\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{2n+1}}=\frac{1}{5}-\frac{1}{\sqrt{2n+1}}.\)

Now let \(n\to\infty\). Since \(\frac{1}{\sqrt{2n+1}}\to 0\),

\(\sum_{k=13}^{\infty}u_k=\frac15.\)

Answer: \(\sum_{k=13}^{n} u_k = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\).

Hence, \(\sum_{k=13}^{\infty} u_k = \frac{1}{5}\).

Write out the sum:

\(\left(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{27}}\right)+\left(\frac{1}{\sqrt{27}}-\frac{1}{\sqrt{29}}\right)+\cdots+\left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right).\)

This is a telescoping series, so all the middle terms cancel, leaving

\(\sum_{k=13}^{n} u_k = \frac{1}{\sqrt{25}}-\frac{1}{\sqrt{2n+1}} = \frac{1}{5}-\frac{1}{\sqrt{2n+1}}\).

Now let \(n \to \infty\). Since \(\frac{1}{\sqrt{2n+1}} \to 0\),

\(\sum_{k=13}^{\infty} u_k = \frac{1}{5}\).

Answer: We first write

\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)

Hence

\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\sum_{r=1}^{N}\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)

Writing out the terms,

\(\dfrac12\left(\dfrac13-\dfrac15\right)+\dfrac12\left(\dfrac15-\dfrac17\right)+\cdots+\dfrac12\left(\dfrac{1}{2N+1}-\dfrac{1}{2N+3}\right).\)

The middle terms cancel, so

\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac13-\dfrac{1}{2N+3}\right)=\dfrac16-\dfrac{1}{2(2N+3)}.\)

Now

\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\sum_{r=1}^{2N}\dfrac{1}{(2r+1)(2r+3)}-\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)

Using the result above,

\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\left(\dfrac16-\dfrac{1}{2(4N+3)}\right)-\left(\dfrac16-\dfrac{1}{2(2N+3)}\right).\)

So

\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2N+3}-\dfrac{1}{4N+3}\right)=\dfrac{N}{(2N+3)(4N+3)}.\)

Since \((2N+3)(4N+3)\gt (2N)(4N)\),

\(\dfrac{N}{(2N+3)(4N+3)}\lt \dfrac{N}{(2N)(4N)}=\dfrac{1}{8N}.\)

Therefore

\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}\lt \dfrac{1}{8N}.\)

Let

\(S_N=\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)

Use partial fractions:

\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac{A}{2r+1}+\dfrac{B}{2r+3}.\)

Then

\(1=A(2r+3)+B(2r+1).\)

Comparing coefficients gives \(2A+2B=0\), so \(B=-A\). Also \(3A+B=1\), hence \(3A-A=1\), so \(2A=1\) and \(A=\dfrac12\). Therefore \(B=-\dfrac12\).

So

\(\dfrac{1}{(2r+1)(2r+3)}=\dfrac12\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)

Hence

\(S_N=\dfrac12\sum_{r=1}^{N}\left(\dfrac{1}{2r+1}-\dfrac{1}{2r+3}\right).\)

Writing out the sum:

\(S_N=\dfrac12\left(\left(\dfrac13-\dfrac15\right)+\left(\dfrac15-\dfrac17\right)+\cdots+\left(\dfrac{1}{2N+1}-\dfrac{1}{2N+3}\right)\right).\)

This is telescoping, so all interior terms cancel and

\(S_N=\dfrac12\left(\dfrac13-\dfrac{1}{2N+3}\right)=\dfrac16-\dfrac{1}{2(2N+3)}.\)

This shows that

\(\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}=\dfrac16-\dfrac{1}{2(2N+3)}.\)

Now let

\(T=\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}.\)

Then

\(T=\sum_{r=1}^{2N}\dfrac{1}{(2r+1)(2r+3)}-\sum_{r=1}^{N}\dfrac{1}{(2r+1)(2r+3)}.\)

Applying the formula for the partial sum gives

\(T=\left(\dfrac16-\dfrac{1}{2(4N+3)}\right)-\left(\dfrac16-\dfrac{1}{2(2N+3)}\right).\)

Therefore

\(T=\dfrac12\left(\dfrac{1}{2N+3}-\dfrac{1}{4N+3}\right).\)

Combining the fractions,

\(T=\dfrac12\cdot\dfrac{(4N+3)-(2N+3)}{(2N+3)(4N+3)}=\dfrac12\cdot\dfrac{2N}{(2N+3)(4N+3)}=\dfrac{N}{(2N+3)(4N+3)}.\)

Now \(2N+3\gt 2N\) and \(4N+3\gt 4N\), so

\((2N+3)(4N+3)\gt (2N)(4N).\)

Since all quantities are positive,

\(T=\dfrac{N}{(2N+3)(4N+3)}\lt \dfrac{N}{(2N)(4N)}=\dfrac{1}{8N}.\)

Hence

\(\sum_{r=N+1}^{2N}\dfrac{1}{(2r+1)(2r+3)}\lt \dfrac{1}{8N}.\)

Answer: (a) \(f(r+1)-f(r)=r(r+1)!-(r-1)r!=r!\bigl(r^2+1\bigr)\).

Hence \(r!\bigl(r^2+1\bigr)=f(r+1)-f(r)\), so the sum telescopes:

\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=\sum_{r=n+1}^{2n}\bigl(f(r+1)-f(r)\bigr)=f(2n+1)-f(n+1)\).

Therefore

\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=2n(2n+1)!-n(n+1)!\).

First simplify \(f(r+1)-f(r)\) using \(f(r)=r!(r-1)\).

We have \(f(r+1)=(r+1)!\cdot r=r(r+1)!\), and \(f(r)=r!(r-1)\). So

\(\begin{aligned} f(r+1)-f(r) &=r(r+1)!-(r-1)r!\\ &=r\cdot(r+1)r!- (r-1)r!\\ &=r!\bigl(r(r+1)-(r-1)\bigr)\\ &=r!\bigl(r^2+r-r+1\bigr)\\ &=r!\bigl(r^2+1\bigr). \end{aligned}\)

So \(r!\bigl(r^2+1\bigr)=f(r+1)-f(r)\), which is ideal for a telescoping sum.

Now

\(\begin{aligned} \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr) &=\sum_{r=n+1}^{2n} \bigl(f(r+1)-f(r)\bigr)\\ &=(f(n+2)-f(n+1))+(f(n+3)-f(n+2))+\cdots +(f(2n+1)-f(2n))\\ &=f(2n+1)-f(n+1). \end{aligned}\)

Finally, substitute into \(f(r)=r!(r-1)\):

\(f(2n+1)=(2n+1)!(2n)=2n(2n+1)!\), and \(f(n+1)=(n+1)!\,n=n(n+1)!\).

Hence

\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=2n(2n+1)!-n(n+1)!\).

Answer: (a) \(S_1=3\), \(S_2=10\), \(S_3=21\), \(S_4=36\).

So \(u_r=4r-1\).

\(\sum_{r=n+1}^{2n}u_r=6n^2+n\).

Since \(S_n=\sum_{r=1}^n u_r=2n^2+n\), substitute \(n=1,2,3,4\):

\(S_1=2(1)^2+1=3\), \(S_2=2(2)^2+2=10\), \(S_3=2(3)^2+3=21\), \(S_4=2(4)^2+4=36\).

The terms are found by taking differences of successive partial sums:

\(u_1=S_1=3\),

\(u_2=S_2-S_1=10-3=7\),

\(u_3=S_3-S_2=21-10=11\),

\(u_4=S_4-S_3=36-21=15\).

These form an arithmetic sequence with first term \(3\) and common difference \(4\). Hence

\(u_r=3+(r-1)4=4r-1\).

Now

\(\sum_{r=n+1}^{2n}u_r=\sum_{r=n+1}^{2n}(4r-1)=4\sum_{r=n+1}^{2n}r-\sum_{r=n+1}^{2n}1\).

There are \(n\) terms from \(r=n+1\) to \(r=2n\), so

\(\sum_{r=n+1}^{2n}r=\frac{n}{2}\big((n+1)+2n\big)=\frac{n(3n+1)}{2}\).

Therefore

\(\sum_{r=n+1}^{2n}(4r-1)=4\cdot \frac{n(3n+1)}{2}-n=2n(3n+1)-n=6n^2+n\).

Answer: 3, 10, 21, 36

\(u_r = 4r - 1\)

\(\sum_{r=n+1}^{2n} u_r = 6n^2 + n\)

First calculate the partial sums from \(S_n=2n^2+n\).

\(S_1=2(1)^2+1=3\)

\(S_2=2(2)^2+2=10\)

\(S_3=2(3)^2+3=21\)

\(S_4=2(4)^2+4=36\)

So the first four sums are \(3,10,21,36\).

The terms of the sequence are found from differences of consecutive sums:

\(u_r=S_r-S_{r-1}\).

Now \(S_r=2r^2+r\), and

\(S_{r-1}=2(r-1)^2+(r-1)\).

Hence

\(u_r=(2r^2+r)-\bigl(2(r-1)^2+(r-1)\bigr)\)

\(=2r^2+r-2(r^2-2r+1)-r+1\)

\(=2r^2+r-2r^2+4r-2-r+1\)

\(=4r-1\).

Therefore \(u_r=4r-1\).

For the required sum,

\(\sum_{r=n+1}^{2n} u_r=\sum_{r=n+1}^{2n}(4r-1)\).

There are \(n\) terms, and they form an arithmetic progression with first term \(u_{n+1}=4(n+1)-1=4n+3\) and last term \(u_{2n}=8n-1\).

So

\(\sum_{r=n+1}^{2n} u_r=\frac{n}{2}\bigl((4n+3)+(8n-1)\bigr)\)

\(=\frac{n}{2}(12n+2)=6n^2+n\).

Answer: \(\displaystyle \frac{1}{r(r+1)(r-1)}=\frac{1}{2(r-1)}-\frac{1}{r}+\frac{1}{2(r+1)}\).

\(\displaystyle \sum_{r=2}^{n}\frac{1}{r(r+1)(r-1)}=\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}\).

\(\displaystyle \sum_{r=2}^{\infty}\frac{1}{r(r+1)(r-1)}=\frac14\).

Write the partial fraction decomposition as

\(\frac{1}{r(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r}+\frac{C}{r+1}.\)

Multiplying by \(r(r-1)(r+1)\) gives

\(1=Ar(r+1)+B(r-1)(r+1)+Cr(r-1).\)

Put \(r=1\): \(1=2A\), so \(A=\frac12\). Put \(r=0\): \(1=-B\), so \(B=-1\). Put \(r=-1\): \(1=2C\), so \(C=\frac12\). Hence

\(\frac{1}{r(r-1)(r+1)}=\frac{1}{2(r-1)}-\frac{1}{r}+\frac{1}{2(r+1)}.\)

Therefore

\(\sum_{r=2}^{n}\frac{1}{r(r-1)(r+1)}=\sum_{r=2}^{n}\left\{\frac{1}{2(r-1)}-\frac1r+\frac{1}{2(r+1)}\right\}.\)

Writing out the first and last terms shows the cancellation:

\(\left(\frac12-\frac12+\frac16\right)+\left(\frac14-\frac13+\frac18\right)+\cdots+\left(\frac{1}{2(n-1)}-\frac1n+\frac{1}{2(n+1)}\right).\)

All middle terms cancel, leaving

\(\sum_{r=2}^{n}\frac{1}{r(r-1)(r+1)}=\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}.\)

As \(n\to\infty\), the terms \(-\frac{1}{2n}\) and \(\frac{1}{2(n+1)}\) tend to \(0\). Thus

\(\sum_{r=2}^{\infty}\frac{1}{r(r-1)(r+1)}=\frac14.\)

Answer: The result is

\(\sum_{n=1}^{N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(N+1)(2N+1)}\)

and also

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\lt \frac{3}{8N^2}.\)

Let

\(H_k:\ \sum_{n=1}^{k} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(k+1)(2k+1)}.\)

We prove \(H_k\) by induction.

First, for \(k=1\),

\(\sum_{n=1}^{1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=\frac{5}{1\cdot 2\cdot 1\cdot 3}=\frac{5}{6},\)

and

\(1-\frac{1}{(1+1)(2\cdot 1+1)}=1-\frac{1}{6}=\frac{5}{6}.\)

So \(H_1\) is true.

Now assume \(H_k\) is true for some positive integer \(k\). Then

\(\sum_{n=1}^{k+1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=1-\frac{1}{(k+1)(2k+1)}+\frac{4k+5}{(k+1)(k+2)(2k+1)(2k+3)}.\)

Write the two terms with a common denominator:

\(\frac{1}{(k+1)(2k+1)}-\frac{4k+5}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{(k+2)(2k+3)-(4k+5)}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{2k^2+3k+1}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{(k+1)(2k+1)}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{1}{(k+2)(2k+3)}.\)

Hence

\(\sum_{n=1}^{k+1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(k+2)(2k+3)},\)

which is exactly \(H_{k+1}\). Therefore \(H_k\) is true for all positive integers \(k\).

Now apply the result with \(k=2N\) and \(k=N\):

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=\sum_{n=1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}-\sum_{n=1}^{N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=\left(1-\frac{1}{(2N+1)(4N+1)}\right)-\left(1-\frac{1}{(N+1)(2N+1)}\right).\)

So

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=\frac{1}{(N+1)(2N+1)}-\frac{1}{(2N+1)(4N+1)}.\)

Factorising gives

\(=\frac{3N}{(N+1)(2N+1)(4N+1)}.\)

Since \(N+1\gt N\), \(2N+1\gt 2N\) and \(4N+1\gt 4N\),

\(\frac{3N}{(N+1)(2N+1)(4N+1)}\lt \frac{3N}{N\cdot 2N\cdot 4N}=\frac{3}{8N^2}.\)

Therefore

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\lt \frac{3}{8N^2}.\)

Answer: \(\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).

Hence

\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+3}\right)=\frac{1}{6}-\frac{1}{2(2n+3)}\).

Therefore

\(\sum_{r=1}^{\infty}\frac{1}{(2r+1)(2r+3)}=\frac{1}{6}\).

Write the term as partial fractions:

\(\frac{1}{(2r+1)(2r+3)}=\frac{A}{2r+1}+\frac{B}{2r+3}\).

So

\(1=A(2r+3)+B(2r+1)\).

Comparing coefficients of \(r\): \(2A+2B=0\), so \(B=-A\). Comparing constants gives \(3A+B=1\), hence \(2A=1\) and \(A=\frac12\), \(B=-\frac12\). Therefore

\(\frac{1}{(2r+1)(2r+3)}=\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).

Now sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac12\sum_{r=1}^{n}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)\).

Writing out the first few terms shows the cancellation:

\(\frac12\left(\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\cdots+\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\right)\).

Everything in the middle cancels, leaving

\(\sum_{r=1}^{n}\frac{1}{(2r+1)(2r+3)}=\frac12\left(\frac13-\frac{1}{2n+3}\right)=\frac16-\frac{1}{2(2n+3)}\).

As \(n\to\infty\), the final term tends to \(0\), so

\(\sum_{r=1}^{\infty}\frac{1}{(2r+1)(2r+3)}=\frac16\).

Answer: (a) \((r+1)^4-r^4=4r^3+6r^2+4r+1\).

(b) \(\displaystyle \sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\).

Expand using the binomial theorem:

\((r+1)^4=r^4+4r^3+6r^2+4r+1\).

Therefore

\((r+1)^4-r^4=4r^3+6r^2+4r+1\).

Now sum this identity from \(r=1\) to \(r=n\):

\(\displaystyle \sum_{r=1}^{n}\big((r+1)^4-r^4\big)=\sum_{r=1}^{n}(4r^3+6r^2+4r+1)\).

The left-hand side telescopes:

\((2^4-1^4)+(3^4-2^4)+\cdots+((n+1)^4-n^4)=(n+1)^4-1\).

Let \(S=\displaystyle \sum_{r=1}^{n}r^3\). Then

\((n+1)^4-1=4S+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n\).

Using

\(\displaystyle \sum_{r=1}^{n}r=\frac{n(n+1)}{2}\), and \(\displaystyle \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}\),

we get

\((n+1)^4-1=4S+n(n+1)(2n+1)+2n(n+1)+n\).

Since \((n+1)^4-1=n^4+4n^3+6n^2+4n\), it follows that

\(4S=n^4+4n^3+6n^2+4n-n(n+1)(2n+1)-2n(n+1)-n\).

Simplifying the right-hand side gives

\(4S=n^4+2n^3+n^2=n^2(n+1)^2\).

Hence

\(\displaystyle \sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\).