MF26

$LHS = \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= \sum_{r=2}^n \frac{1}{2} (\frac{1}{r-1} - \frac{1}{r+1})$
$= \frac{1}{2} \sum_{r=2}^n \frac{1}{r-1} - \frac{1}{r+1}$
$= \frac{1}{2} [\frac{1}{1} - \frac{1}{3}$
$+ \frac{1}{2} - \frac{1}{4}$
$+ \frac{1}{3} - \frac{1}{5}$

$+ \frac{1}{n-3} - \frac{1}{n-1}$
$+ \frac{1}{n-2} - \frac{1}{n}$
$+ \frac{1}{n-1} - \frac{1}{n+1}]$
$= \frac{1}{2} [\frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}]$
$= \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$

Since the $lim_{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= lim_{n \rightarrow \infty} \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$
$= \frac{3}{4} - 0 - 0$
$- \frac{3}{4}$ is a constant, the series convergences.
The sum to infinity $= \frac{3}{4}$

Let $T_n$ denote the $n^{th}$ term of the AP.
$T_n = a + (n-1)d$
$T_1 = a$
$T_3 = a +2d$
$T_6 = a +5d$
Since they are consecutive terms of a GP,
$\frac{T_3}{T_1} = \frac{T_6}{T_3} = r$
$\Rightarrow \frac{a+2d}{a} = \frac{a+5d}{a+2d}$
$(a+2d)^2 = a(a+5d)$
$a^2 + 4ad + 4d^2 = a^2 + 5ad$
$4d^2 - ad =0$
$d(4d - a) = 0$
$d = 0 (NA) \mathrm{~or~} 4d = a$
$\Rightarrow r = \frac{a+2d}{a}$
$r = \frac{6d+2d}{4d} = \frac{3}{2} > 1$, thus its not convergent

$S_{15} = \frac{n}{2}(2a + (n-1)d)$
$= \frac{15}{2}(2a + 14 (\frac{a}{4}))$
$= 41.25 a$

Sum of first 3 terms $= \frac{3}{2} (2a+(3-1)d)$
$6 = 3a+3d$
$a+d=2$ —(1)
Sum of last 3 terms $= \frac{3}{2}[2(a+(n-1)d) + (3-1)(-d)]$; Here we consider an AP that has first term $T_n = a + (n-1)d$ and common difference $-d$.
$\Rightarrow 231 = \frac{3}{2}(2a + 2nd - 2d -2d)$
$231 = 3a + 3nd - 6d$
$77 = a + nd -2d$ —(2)
Sum of n terms = $\frac{n}{2}[2a+ (n-1)d]$
$1106 = \frac{n}{2}(2a + nd - d)$ —(3)
Solve for n.

(i) Volume, $V = \pi r^2 h$
Volume of kth later, $V_k = \pi [(20)(\frac{5}{6})^{k-1}]^2(22)(\frac{4}{5})^{k-1}$
$V_k = 8800 \pi [\frac{25}{36} \times \frac{4}{5}]^{k-1}$
$V_k = 8800 \pi (\frac{5}{9})^{k-1}$
(ii)
Since $latex r = \frac{5}{9} <1, S_{\infty}$ exists. Theoretical Max Volume, $latex S_{\infty} = \frac{8800 \pi}{1 - \frac{5}{9}} = 19800 \pi$. Total Volume, $latex S_n = \frac{8800 \pi (1 - (\frac{5}{9})^n)}{1 - \frac{5}{9}}$ We want $latex S_n \le 0.95 S_{\infty}$

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