Question

The sum of the series $4+8+16+32+.......$. till $10$ terms is

• A. $6\times 2^{10}-1$
• B. $4\times 2^{10}+1$
• C. $4(2^{10}-1)$
• D. $4(2^{10}+1)$

Answer 1

The correct option is C $4(2^{10}-1)$

Let $S_{10}=4+8+16+32+........$ upto $10$ terms
$=4(1+2+4+8+.........$.upto $10$ terms
Clearly, $1+2+2^2+2^3+...$ is an GP with first term $=1$ and common ratio $=2$.
$\therefore S_{10}=4[\frac{1(2^{10}-1)}{2-1}]=4(2^{10}-1)$

Thus the sum of the given series is $4(2^{10}-1)$