Question

The $n^{\text{th}}$ term of the series
$1+4+13+40+121+364+...$ is :

• A. $3^n-1$
• B. $\frac{1}{2}(3^n+1)$
• C. $\frac{1}{2}(3^n-1)$
• D. $\left(\frac{2^n+1}{2}\right)$

Answer 1

The correct option is C $\frac{1}{2}(3^n-1)$

Let $S_n=1+4+13+40+121+364+...+T_n$
$S_n=(1+4+13+40+121+...+T_{n-1})+T_n$
Subtracting the above equations, we get
$0=1+3+9+27+81+...+(T_n-T_{n-1})-T_n\Rightarrow T_n=1+3+9+27+81+...+(T_n-T_{n-1})\text{upto n terms}$

$\Rightarrow T_n=\frac{3^n-1}{3-1}\Rightarrow T_n=\frac{3^n-1}{2}$