Question

For a sequence $\\{a_n\\},a_1=2$ and $\frac{n+1}{a_n}=\frac{1}{3}$. Then $\displaystyle\sum_{r=1}^{20}a_r$ is

• A. $\frac{20}{2}[4+19\times3]$
• B. $3\left(1-\frac{1}{3^{20}}\right)$
• C. $2(1-3^{20})$
• D. none of these.

Answer 1

Answer: (B)

$3\left(1-\frac{1}{3^{20}}\right)$

Answer 2

$\frac{a_{n+1}}{a_{n}} = \frac{1}{3}$ $\Rightarrow$ Common ratio $= \frac{1}{3} = R$ (say) First term $= a_{1}=2$ $\sum\limits_{r=0}^{20} a_{r} = \frac{a_{1}\left[1-R^{20}\right]}{1-R} = \frac{2\left[1-\left(\frac{1}{3}\right)^{20}\right]}{1-\frac{1}{3}}$ $= \frac{2\left[1-\frac{1}{3^{20}}\right]}{\frac{2}{3}} = 3\left[1-\frac{1}{3^{20}}\right]$