If the sum of the first n terms of the seriesn sqrt(3) + sqrt(75) + sqrt(243) + sqrt(507) + ... isn 435sqrt(3), then n equals

\ (c): We have, sqrt(3)+sqrt(75)+sqrt(243)+ * s ... =435 sqrt(3) => sqrt(3)[1+sqrt(25)+sqrt(81)+sqrt(169)+ * s ... ]=.435 sqrt(3) => sqrt(3)[1+5+9+13+ * s ... ]=.435 sqrt(3) => sqrt(3) x (n / 2)[2+(n-1) 4]=435 sqrt(3) [Because 1+5+9+13 ... . . is an A.P. ] => n+(4 n^2 / 2)-(4 n / 2)=435 => 2 n+4 n^2-4 n=870 => 4 n^2-2 n-870=0 => 2 n^2-n-435=0 => n=frac1 ± sqrt(1+4 x 2 x 435)4=(1 ± 59 / 4) => n=(1+59 / 4)=15[Because (1-59 / 4) is neglected ]\

If the sum of the first n terms of the series \sqrt{3... | Sequence & Series (Mathematics)

\begin{aligned}& \text { (c): We have, } \sqrt{3}+\sqrt{75}+\sqrt{243}+\cdots \ldots=435 \sqrt{3} \\& \Rightarrow \sqrt{3}[1+\sqrt{25}+\sqrt{81}+\sqrt{169}+\cdots \ldots]=.435 \sqrt{3} \\& \Rightarrow \sqrt{3}[1+5+9+13+\cdots \ldots]=.435 \sqrt{3} \\& \Rightarrow \sqrt{3} \times \frac{n}{2}[2+(n-1) 4]=435 \sqrt{3} \\& {[\because 1+5+9+13 \ldots . . \text { is an A.P. }]} \\& \Rightarrow n+\frac{4 n^2}{2}-\frac{4 n}{2}=435 \Rightarrow 2 n+4 n^2-4 n=870 \\& \Rightarrow 4 n^2-2 n-870=0 \Rightarrow 2 n^2-n-435=0 \\& \Rightarrow n=\frac{1 \pm \sqrt{1+4 \times 2 \times 435}}{4}=\frac{1 \pm 59}{4} \\& \Rightarrow n=\frac{1+59}{4}=15\left[\because \frac{1-59}{4} \text { is neglected }\right]\end{aligned}

Explanation

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