If the sum of the first n terms of the series | vedclass

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(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3}

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$15$

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(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3} + \dots$Factor out $\sqrt{3}$: $\sqrt{3}(1 + 5 + 9 + 13 + \dots + T_n) = 435\sqrt{3}$Dividing both sides by $\sqrt{3}$,we get the sum of an arithmetic progression: $1 + 5 + 9 + 13 + \dots + T_n = 435$Here,the first term $a = 1$ and the common difference $d = 4$.The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values: $\frac{n}{2}[2(1) + (n - 1)4] = 435$$\frac{n}{2}[2 + 4n - 4] = 435$$\frac{n}{2}[4n - 2] = 435$$n(2n - 1) = 435$$2n^2 - n - 435 = 0$Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{1 \pm \sqrt{(-1)^2 - 4(2)(-435)}}{2(2)} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4} = \frac{1 \pm 59}{4}$Since $n$ must be positive,$n = \frac{1 + 59}{4} = \frac{60}{4} = 15$.

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

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