Question

Find the sum of n terms of the series $\frac{4}{\mathrm{1.2.3}}\cdot \frac{1}{3}+\frac{5}{\mathrm{2.3.4}}\cdot \frac{1}{3^2}+\frac{6}{\mathrm{3.4.5}}\cdot \frac{1}{3^3}+....$

Answer 1

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

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

$u_n=\frac{1}{2n(n+1)}\cdot \frac{1}{3^{n-1}}-\frac{1}{2(n+1)(n+2)}\cdot \frac{1}{3^n}$

$u_{n-1}=\frac{1}{2(n-1)n}\cdot \frac{1}{3^{n-2}}-\frac{1}{2\left(n\right)(n+1)}\cdot \frac{1}{3^{n-1}}$

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$u_1=\frac{1}{2\cdot 2}\cdot \frac{1}{3^0}-\frac{1}{2\cdot 2\cdot 3}\cdot \frac{1}{3^1}$

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