9231 P11 - Nov 2020 - Q2 - 8 marks
5802
2 (a) Use standard results from the List of Formulae (MF19) to show that
(b) Use the method of differences to find \(\sum_{r=1}^{n} \frac{1}{(7 r+1)(7 r+8)}\) in terms of \(n\).
(c) Deduce the value of \(\sum_{r=1}^{\infty} \frac{1}{(7 r+1)(7 r+8)}\).
Solution
Mark Scheme
Solution
Checked by expert
Answer:
\(a=\frac{49}{3}\), \(b=56\), \(c=\frac{143}{3}\), so \(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+56n^2+\frac{143}{3}n\). \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\). \(\sum_{r=1}^{\infty}\frac{1}{(7r+1)(7r+8)}=\frac{1}{56}\). Expand the product inside the sum:
\((7r+1)(7r+8)=49r^2+56r+7r+8=49r^2+63r+8\).
So
\(\sum_{r=1}^{n}(7r+1)(7r+8)=\sum_{r=1}^{n}(49r^2+63r+8)\).
Using \(\sum_{r=1}^{n}r^2=\frac{1}{6}n(n+1)(2n+1)\), \(\sum_{r=1}^{n}r=\frac{1}{2}n(n+1)\), and \(\sum_{r=1}^{n}1=n\),
\(\sum_{r=1}^{n}(49r^2+63r+8)=49\left(\frac{1}{6}n(n+1)(2n+1)\right)+63\left(\frac{1}{2}n(n+1)\right)+8n\).
Expanding gives
\(49\left(\frac{1}{6}n(n+1)(2n+1)\right)=\frac{49}{3}n^3+\frac{49}{2}n^2+\frac{49}{6}n\),
and
\(63\left(\frac{1}{2}n(n+1)\right)=\frac{63}{2}n^2+\frac{63}{2}n\).
Therefore
\(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+\left(\frac{49}{2}+\frac{63}{2}\right)n^2+\left(\frac{49}{6}+\frac{63}{2}+8\right)n\).
Hence
\(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+56n^2+\frac{143}{3}n\).
So \(a=\frac{49}{3}\), \(b=56\), and \(c=\frac{143}{3}\).
First write the fraction in partial fractions:
\(\frac{1}{(7r+1)(7r+8)}=\frac{A}{7r+1}+\frac{B}{7r+8}\).
Multiplying through by \((7r+1)(7r+8)\),
\(1=A(7r+8)+B(7r+1)\).
Putting \(r=-\frac{1}{7}\) gives \(1=7A\), so \(A=\frac{1}{7}\). Putting \(r=-\frac{8}{7}\) gives \(1=-7B\), so \(B=-\frac{1}{7}\).
Thus
\(\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{7r+1}-\frac{1}{7r+8}\right)\).
Now sum from \(r=1\) to \(n\):
\(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\sum_{r=1}^{n}\left(\frac{1}{7r+1}-\frac{1}{7r+8}\right)\).
Writing out the terms,
\(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+\frac{1}{22}-\frac{1}{29}+\cdots+\frac{1}{7n+1}-\frac{1}{7n+8}\right)\).
The intermediate terms cancel, leaving
\(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\).
From part (b),
\(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\).
As \(n\to\infty\), \(\frac{1}{7n+8}\to 0\). Therefore
\(\sum_{r=1}^{\infty}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\cdot\frac{1}{8}=\frac{1}{56}\).
