Answer :

In the given question we need to find the sum of the series.

For that, first, we need to find the n^{th} term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is 3, 15, 35, 63 … to n terms.

The series can be written as, [2^{2} - 1, 4^{2}-1, 6^{2}-1… (2n)^{2}-1].

So, n^{th} term of the series,

a_{n} = (2n)^{2} - 1

a_{n} = 4n^{2} - 1

Now, we need to find the sum of this series, S_{n.}

__Note:__

I. Sum of first n natural numbers, 1 + 2 +3+…n,

II. Sum of squares of first n natural numbers, 1^{2} + 2^{2} + 3^{2}+….n^{2},

III. Sum of cubes of first n natural numbers, 1^{3} + 2^{3} + 3^{3} +…..n^{3},

IV. Sum of a constant k, N times,

So, for the given series, we need to find,

From, the above identities,

So, Sum of the series,

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