Solution :
By using method of differences,
The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. 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 + 7 + 14 + 24 + 37 + ….. + \(T_{n-1}\) + \(T_n\) ……(i)
Also, \(S_n\) = 3 + 7 + 14 + 24 + 37 + ….. + \(T_{n-1}\) + \(T_n\) ……(ii)
Subtracting (ii) from (i), we get
0 = 3 + [4 + 7 + 10 + 13 + …… + \(T_{n-1}\) + \(T_n\)] – \(T_n\)
\(\implies\) \(T_n\) = 3 + \((n-1)\over 2\){2*4+(n-1-1)*3} = \(6 + (n-1)(3n+2)\over 4\)
\(\implies\) \(T_n\) = \(1\over 2\) (\(3n^2 – n + 4\))
\(\therefore\) \(S_n\) = \(\sum_{k=1}^{n}\) \(T_k\) = \(\sum_{k=1}^{n}\) \(1\over 2\) (\(3n^2 – n + 4\)) = \(1\over 2\)[\(\sum_{k=1}^{n}\)\(3k^2\) – \(\sum_{k=1}^{n}\) k + 4n]
\(\implies\) \(S_n\) = \(1\over 2\)[3\(n(n+1)(2n+1)\over 6\) – \(n(n+1)\over 2\) + 4n] = \(n\over 2\) (\(n^2 + n + 4\))
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