Question

Sum of the series $11+23+45+87...$ up to $n$ terms

• A. $10(2^n-1)+n^2$
• B. $10(2^n+1)+n^2$
• C. $20(2^n-n)+n^2$
• D. None of these

Answer 1

Answer: (A)

$10(2^n-1)+n^2$

Let $S=11+23+45+87+...$up to $n$ terms
$\Rightarrow S=(5.2+1)+(5.2^2+3)+(5.2^3+5)+(5.2^4+7)+...$up to $n$ terms
Now, $a_n=5{.2}^n+2n-1$
Therefore, $S={\sum }_{n=1}^n{a}_n={\sum }_{n=1}^n(5{.2}^n+2n-1)$
$\Rightarrow S=5{\sum }_{n=1}^n{2}^n+2{\sum }_{n=1}^{n}n-{\sum }_{n=1}^{n}1=\frac{5.2(2^n-1)}{2-1}+\frac{2n(n+1)}{2}-n$
$\Rightarrow S=10(2^n-1)+n^2$