Question

Sum of the series $3 + 5 + 9 + 17 + 33 + ...$ to $n$ terms is

• A. $2^{n + 1} - n - 2$
• B. $2^{n + 1} + n -2$
• C. $2^{n} + n - 2$
• D. $2^{n + 1} - n + 2$

Answer 1

Answer: (B)

Given terms are $3, 5, 9, 17, 33, ....$
Series formed from the differences of the given series is $2, 4, 8, 16, ... \in G.P$.
$\therefore ... t_{n} = A 2^{n -1} + B$
$\Rightarrow t_{1}= 3, t_{2} = 5$ (Using $3 = A + B$ and $5 = 2A + B$
$\Rightarrow A = 2 , B = 1$)
$\therefore t_{n} = 2^{n} +1$
$\therefore S_{n} = \Sigma t_{n} = \Sigma\left(2^{n} + 1\right)$
$= \Sigma2^{n} +\Sigma1 = 2 \left(2^{n} -1\right) + n = 2^{n + 1} + n - 2$
Short Cut Method : $S_{n} = 3 + 5 + 9 + 17 + 33 + .....$
$= \left( 2 +1\right) +\left(2^{2} +1\right) +\left(2^{3} +1\right) +\left(2^{4} +1\right) + ......$
$= \left(2 +2^{2} +2^{3} +2^{4} +...n \, terms \right) +n$
$= 2 \left(2^{n} -1\right) +n = 2^{n +1} +n -2$