Answer: (a) \(2n^4+8n^3+10n^2+7n\)

(b) \(1-\dfrac1{(n+1)^4}\)

(c) \(1\)

(a)

\(\sum_{r=1}^{n}(8r^3+12r^2+4r+3)=8\sum r^3+12\sum r^2+4\sum r+3n.\)

Using MF19,

\(8\left(\frac{n(n+1)}2\right)^2+12\left(\frac{n(n+1)(2n+1)}6\right)+4\left(\frac{n(n+1)}2\right)+3n\)

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

Expanding and simplifying gives

\(2n^4+8n^3+10n^2+7n.\)

(b)

First put the left-hand side over a common denominator:

\(\frac1{r^4}-\frac1{(r+1)^4}=\frac{(r+1)^4-r^4}{r^4(r+1)^4}.\)

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

\(\frac1{r^4}-\frac1{(r+1)^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}.\)

Therefore

\(\sum_{r=1}^{n}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}=\sum_{r=1}^{n}\left(\frac1{r^4}-\frac1{(r+1)^4}\right).\)

The terms telescope:

\(1-\frac1{2^4}+\frac1{2^4}-\frac1{3^4}+\cdots+\frac1{n^4}-\frac1{(n+1)^4}=1-\frac1{(n+1)^4}.\)

(c)

From part (b), the partial sum is \(1-\frac1{(n+1)^4}\). Hence

\(\lim_{n\to\infty}\left(1-\frac1{(n+1)^4}\right)=1.\)

Mark scheme