If the sum of the first n terms of the series √3+√75+√243+√507+ is 435√3, then n equals:

Question

If the sum of the first $n$ terms of the series $\sqrt3+\sqrt{75}+\sqrt{243}+\sqrt{507}+\cdots$ is $435\sqrt3$, then $n$ equals:

Accepted Answer

Given,

$$\sqrt 3 $$  + $$\sqrt {75} $$ + $$\sqrt {243} $$ + $$\sqrt {507} $$ + . . . . . .+ n terms

= $$\sqrt 3 $$ + $$\sqrt {25 \times 3} $$ + $$\sqrt {81 \times 3} $$ + $$\sqrt {169 \times 3} $$ + . . . . . .+ n terms

= $$\sqrt 3 $$ + 5$$\sqrt 3 $$ + 9$$\sqrt 3 $$ + 13$$\sqrt 3 $$ + . . . . . .+ n terms

= $$\sqrt 3 $$ [ 1 + 5 + 9 + 13 + . . . . .+ n  terms]

= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {2.1 + \left( {n - 1} \right)4} \right)} \right]$$

= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {2 + 4n - 4} \right)} \right]$$

= $$\sqrt 3 $$ $$\left[ {{n \over 2}\left( {4n - 2} \right)} \right]$$

= $$\sqrt 3 $$ [n (2n - 1)]

According to question,

$$\sqrt 3 $$ [n (2n - 1)] = 435$$\sqrt 3 $$

$$ \Rightarrow $$ $\,\,\,$2n2$-$n = 435$\therefore\,\,\,$n =${{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4}$=${{1 \pm 59} \over 4}$$\therefore\,\,\,$n =${{1 + 59} \over 4}$= 15   or${{1 - 59} \over 4}$=$-$14.5$\therefore\,\,\,$n = 15 (as    n  can't be$-$

Solutions · 1