Find the sum to n terms of the series : 3 + 15 + 35 + 63 + …..

Solution :

By using method of differences,

The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. Clearly,these differences are in AP.

Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series

Then, \(S_n\) = 3 + 15 + 35 + 63 + ….. + \(T_{n-1}\) + \(T_n\) ……(i)

Also, \(S_n\) =         3 + 15 + 35 + ….. + \(T_{n-1}\) + \(T_n\) ……(ii)

Subtracting (ii) from (i), we get

0 = 3 + [12 + 20 + 28 + …… + \(T_{n-1}\) + \(T_n\)] – \(T_n\)

\(\implies\) \(T_n\) = 3 + \((n-1)\over 2\){2*12+(n-1-1)*8} = 3 + (n-1){12+4n-8}

\(\implies\) \(T_n\) = 3 + (n-1)(4n+4) = \(4n^2 – 1\)

\(\therefore\) \(S_n\) = \(\sum_{k=1}^{n}\) \(T_k\) = \(\sum_{k=1}^{n}\) (\(4k^2 – 1\)) = 4\(\sum_{k=1}^{n}\)\(k^2\) – \(\sum_{k=1}^{n}\) 1

\(\implies\) \(S_n\) = 4{\(n(n+1)(2n+1)\over 6\)} – n = \(n\over 3\) (\(4n^2 + 6n – 1\))


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