RI/2021/JC2/Prelim/P1/Q04

(a) Show that $(2 r+1)^3-(2 r-1)^3=12 k r^2+k$, where $k$ is a constant to be determined. Use this result to find $\sum_{r=1}^n r^2$, giving your answer in the form $p n(q n+1)(2 q n+1)$ where $p$ and $q$ are constants to be determined.

[5]

(b) Raabe’s test states that a series of positive terms of the form $\sum_{r=1}^{\infty} a_r$ converges when

and diverges when

When

the test is inconclusive. Using the test, explain why the series $\sum_{r=1}^{\infty} \frac{1}{r^3}$ converges.

[3]

(b)

Let $a_n=\frac{1}{n^3}$.
$n\left(\frac{a_n}{a_{n+1}}-1\right) =n\left[\frac{\left(\frac{1}{n^3}\right)}{\left(\frac{1}{(n+1)^3}\right)}-1\right] $
$=n\left[\frac{(n+1)^3}{n^3}-1\right] $
$=n\left(\frac{n^3+3 n^2+3 n+1-n^3}{n^3}\right) $
$=\left(\frac{3 n^2+3 n+1}{n^2}\right) $
$=3+\frac{3}{n}+\frac{1}{n^2} $

Therefore $\sum_{r=1}^{\infty} \frac{1}{r^3}$ converges.

(a) $(2 r+1)^3-(2 r-1)^3$
$=\left[8 r^3+12 r^2+6 r+1\right]-\left[8 r^3-12 r^2+6 r-1\right]$
$=24 r^2+2$
$\therefore k=2$
OR
$(2 r+1)^3-(2 r-1)^3$
$=[(2 r+1)-(2 r-1)]\left[(2 r+1)^2+(2 r+1)(2 r-1)+(2 r-1)^2\right]$
$=2\left[4 r^2+4 r+1+4 r^2-1+4 r^2-4 r+1\right]$
$=2\left(12 r^2+1\right)$
$=24 r^2+2$
$\therefore k=2$

$\sum_{r=1}^n\left(24 r^2+2\right)=\sum_{r=1}^n\left((2 r+1)^3-(2 r-1)^3\right)$

$24 \sum_{r=1}^n r^2+2 n =(2 n+1)^3-1 $

$sum_{r=1}^n r^2 =\frac{1}{24}\left[(2 n+1)^3-(2 n+1)\right] $

$=\frac{(2 n+1)}{24}\left[(2 n+1)^2-1\right] $

$=\frac{(2 n+1)}{24}(2 n+2)(2 n) $

$=\frac{1}{6} n(n+1)(2 n+1) $

$ \therefore p=\frac{1}{6}, \quad q=1$

(b)

Therefore $\sum_{r=1}^{\infty} \frac{1}{r^3}$ converges.