CAIE A-Level Further Math 1.3.2 Method of differences

Practise finite telescoping sums by rewriting general terms, displaying cancellation and keeping the surviving end terms in n.

Exam points

express rational terms using partial fractions before applying the method of differences
show enough consecutive and final terms to justify the telescoping cancellation

[Maximum number: 5]

Show that

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

and hence use the method of differences to find $\sum_{r=1}^{n} \frac{3 r^{2}+3 r+1}{r^{3}(r+1)^{3}}$ .

$\frac{1}{r^{3}}-\frac{1}{(r+1)^{3}}=\frac{(r+1)^{3}-r^{3}}{r^{3}(r+1)^{3}}=\frac{r^{3}+3 r^{2}+3 r+1-r^{3}}{r^{3}(r+1)^{3}}=\frac{3 r^{2}+3 r+1}{r^{3}(r+1)^{3}}$

M1 A1

Puts over a common denominator and
expands, AG.

$\begin{aligned} \sum_{r=1}^{n} \frac{3 r^{2}+3 r+1}{r^{3}(r+1)^{3}}=\sum_{r=1}^{n}\left(\frac{1}{r^{3}}-\frac{1}{(r+1)^{3}}\right) =1-\frac{1}{2^{3}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\ldots+\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}} \end{aligned}$

M1 A1

Shows three complete terms, including
last.

$1-\frac{1}{(n+1)^{3}}$

A1

5

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1.3.2—Method of differences question 1