If the sum to n terms of the series S=3+75+243+507+⋯ is 4353,then n is equal to

S=3(1)+352+392+3132+⋯=3[1+5+9+13+⋯ upto n terms ]=3n2(2(1)+(n−1)(4))∴ 4353=3n(2n−1)⇒ 2n2−n−435=0⇒ (2n+29)(n−15)=0⇒ n=15 as n∈N

If the sum to n terms of the series S=3+75+243+507+⋯ is 4353,then n is equal to

S=3(1)+352+392+3132+⋯=3[1+5+9+13+⋯ upto n terms ]=3n2(2(1)+(n−1)(4))∴ 4353=3n(2n−1)⇒ 2n2−n−435=0⇒ (2n+29)(n−15)=0⇒ n=15 as n∈N

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If the sum to n terms of the series $S = \sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots is 435 \sqrt{3},$ then n is equal to

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answer is B.

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Detailed Solution

$\begin{array}{l} S = \sqrt{3} (1) + \sqrt{3} \sqrt{5^{2}} + \sqrt{3} \sqrt{9^{2}} + \sqrt{3} \sqrt{13^{2}} + \dots \\ = \sqrt{3} [1 + 5 + 9 + 13 + \dots upto n terms] \\ = \sqrt{3} (\frac{n}{2}) (2 (1) + (n - 1) (4)) \end{array}$

$\begin{aligned} ∴ 435 \sqrt{3} = \sqrt{3} n (2 n - 1) \\ \Rightarrow 2 n^{2} - n - 435 = 0 \\ \Rightarrow (2 n + 29) (n - 15) = 0 \\ \Rightarrow n = 15 as n \in N \end{aligned}$