Question

Sum to $2n$ terms of the series $\frac{5}{7}+\frac{12}{7}+\frac{47}{49}+\frac{96}{49}+...$ up to $2n$ terms

• A. $2n-\frac{2}{3}(1-\frac{1}{7^n})$
• B. $3n-\frac{2}{3}(1-\frac{1}{7^n})$
• C. $n-\frac{2}{3}(1-\frac{1}{7^n})$
• D. $2n-\frac{2}{3}(1-\frac{1}{7^{2n}})$

Answer 1

The correct option is B $3n-\frac{2}{3}(1-\frac{1}{7^n})$

$S=\frac{5}{7}+\frac{12}{7}+\frac{47}{49}+\frac{96}{49}+...$ up to $2n$ terms
$\Rightarrow S=\left(\frac{5+12}{7}\right)+\left(\frac{47+96}{49}\right)+...$ up to $n$ terms
$\Rightarrow S=\left(\frac{7^1-2+2.7^1-2}{7^1}\right)+\left(\frac{7^2-2+2.7^2-2}{7^2}\right)+...$ up to $n$ terms
Now, $a_n=\frac{7^n-2+2.7^n-2}{7^n}=3-\frac{4}{7^n}$
$S={\sum }_{n=1}^n(3-\frac{4}{7^n})={\sum }_{n=1}^{n}3-{\sum }_{n=1}^n\left(\frac{4}{7^n}\right)$
$\Rightarrow S=3n-\left(\frac{4}{7}\right)\left[\frac{1-{\left(1/7\right)}^n}{1-\left(1/7\right)}\right]$
$\Rightarrow S=3n-\frac{2}{3}(1-\frac{1}{7^n})$
Ans: B