Question

Sum of $n$ terms of series $12+16+24+40+\cdots$will be

• A. $2\left(2^n-1\right)+8n$
• B. $2\left(2^n-1\right)+6n$
• C. $3\left(2^n-1\right)+8n$
• D. $4\left(2^n-1\right)+8n$

Answer 1

The correct option is D

$4\left(2^n-1\right)+8n$

Explanation for the correct option :

Step-1 : Reforming the given series

Let $S$ be the sum of the first $n$ terms of the given series $12+16+24+40+\cdots$

i.e. $S=12+16+24+40+\cdots +t_n$, where $t_n$ is the $n^{th}$ term of the series. To find $t_n$ we shall rewrite the series as follows :

$$\begin{gathered} 12+16+24+40+\cdots \\ =\left(8+2^2\right)+\left(8+2^3\right)+\left(8+2^4\right)+\left(8+2^5\right)+\cdots \end{gathered}$$

So, using the above pattern, we get $t_n=8+2^n$.

Step-2 : Finding the value of $S$

We have

$$\begin{gathered} S=\left(8+2^2\right)+\left(8+2^3\right)+\left(8+2^4\right)+\left(8+2^5\right)+\cdots +\left(8+2^n\right) \\ =\left(8+8+\cdots +8\right)+\left(2^2+2^3+\cdots +2^n\right) \\ =8n+\frac{2^2\left(2^{n-1}-1\right)}{2-1} \\ =8n+4\left(2^n-1\right) \end{gathered}$$

Here, we have used the fact that the sum of the first $n$ terms a geometric series with the first term $a$ and common ratio $r$ is $\frac{a\left(r^n-1\right)}{r-1}$.

Hence, option (D) is the correct answer.