CAIE A-Level Further Mathematics 1.3 Summation of Series

Practise applying MF19 power sums, decomposing rational terms and displaying cancellation clearly before simplifying finite sums or taking a justified infinite limit.

[Maximum number: 9]

(a)

Use standard results from the list of formulae (MF19) to find $\sum_{r=1}^{n}\left(3 r^{2}+3 r+1\right)$ in terms of n, simplifying your answer.

[ 3 ]

$\frac{1}{2} n(n+1)(2 n+1)+\frac{3}{2} n(n+1)+n$

M1 A1

Substitutes correct formulae from
MF19.

$n^{3}+3 n^{2}+3 n$

A1

Simplifies

3

(b)

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

[ 5 ]

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

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

(c)

Deduce the value of $\sum_{r=1}^{\infty} \frac{3 r^{2}+3 r+1}{r^{3}(r+1)^{3}}$ .

[ 1 ]

All question bank results loaded

CAIE A-Level Further Mathematics 1.3 Summation of Series

Exam points

Covered syllabus points

Covered syllabus points

1.3 Summation of series question 1

Question (a)

Question (b)

Question (c)