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$The sum of the first n-terms of the series \sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots is 435 \sqrt{3} then n is$

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Correct option is B

given that 31+5+9+13+…+Tn=4353⇒n2[2+4(n−1)]=435⇒2n2−n−435=0 n=15

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The sum of the first n-terms of the series 3+75+243+507+… is 4353 then n is

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$The sum of the first n-terms of the series \sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots is 435 \sqrt{3} then n is$