Question

Find the sum of n terms of the series $\frac{3}{1.2}\cdot \frac{1}{2}+\frac{4}{2.3}\cdot \frac{1}{2^2}+\frac{5}{3.4}\cdot \frac{1}{2^3}+\frac{6}{4.5}\cdot \frac{1}{2^4}+....$

Answer 1

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

Assume

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

$A=2,B=-1$

$\therefore u_n=(\frac{2}{n}-\frac{1}{n+1})\frac{1}{2^n}$

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

$\therefore u_n=4^n-\frac{4^{n+1}}{n+2}+\frac{4^n}{n+1}$

$u_{n-1}=4^{n-1}-\frac{4^n}{n+1}+\frac{4^{n-1}}{n}$

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$u_2=4^2-\frac{1}{4}.4^3+\frac{1}{3}.4^2$

$u_1=4-\frac{1}{3}.4^2+\frac{1}{2}.4$

$S_n=(4+4^2+4^3+....+4^n)-\frac{4^{n+1}}{n+2}+2$

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

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