Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……

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)}

= \(1\over 8\)(\(V_n\) – \(V_{n-1}\))

\(S_n\) = \({\sum}_{r=1}^{n‎} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\))

\(\therefore\)  \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\)

= \(n(2n^3 + 8n^2 + 7n – 2)\)


Similar Questions

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….

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

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is

Leave a Comment

Your email address will not be published.