Solution :
By using method of differences,
The \(n^{th}\) term is (2n-1)(2n+1)(2n+3)
\(T_n\) = (2n-1)(2n+1)(2n+3)
\(T_n\) = \(1\over 8\)(2n-1)(2n+1)(2n+3){(2n+5) – (2n-3)}
= \(1\over 8\)(\(V_n\) – \(V_{n-1}\))
\(S_n\) = \({\sum}_{r=1}^{n} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\))
\(\therefore\) \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\)
= \(n(2n^3 + 8n^2 + 7n – 2)\)
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