Question

Let $a_n=\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$ then $\sum _{n=1}^{144}{a}_n$ equals

• A. $2456$
• B. $2645$
• C. $2466$
• D. $2546$

Answer 1

The correct option is A $2456$

Given $a_n=\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$

Multiplying Dividing by $\sqrt{2n+1}-\sqrt{2n-1}$

$a_n=\frac{(4n+\sqrt{4n^2-1})(\sqrt{2n+1}-\sqrt{2n-1})}{(2n+1)-(2n-1)}$

$a_n=\frac{4n\sqrt{2n+1}-4n\sqrt{2n-1}+(2n+1)\sqrt{2n-1}-(2n-1)\sqrt{2n+1}}{2}$

$a_n=\frac{(2n+1)\sqrt{2n+1}-(2n-1)\sqrt{2n-1}}{2}$

$a_n=\frac{{(2n+1)}^{3/2}-{(2n-1)}^{3/2}}{2}$

$a_1=\frac{{\left(3\right)}^{3/2}-{\left(1\right)}^{3/2}}{2}$

$a_2=\frac{{\left(5\right)}^{3/2}-{\left(3\right)}^{3/2}}{2}$

$a_3=\frac{{\left(7\right)}^{3/2}-{\left(5\right)}^{3/2}}{2}$

$a_n=\frac{{(2n+1)}^{3/2}-{(2n-1)}^{3/2}}{2}$

${\sum }_{i=1}^n{a}_n=\frac{{(2n+1)}^{3/2}-1}{2}$

${\sum }_{i=1}^{144}{a}_n$$=\frac{{(2\times 144+1)}^{3/2}-1}{2}=\frac{{\left(289\right)}^{3/2}-1}{2}=\frac{{\left(17\right)}^3-1}{2}=2456$