Find the sum of the series 1.22 + 3.32 + 5.42 + .... to n terms(a) \(

1.	Find the sum of the series 1.22 + 3.32 + 5.42 + .... to n terms(a) \(\frac{n}{2}\)(n3 + 4n2 + 4n - 1)(b)\(\frac{n}{2}\)(n + 1)2 (2n + 1) (c) \(\frac{n^2}{3}\)(n + 1)2 (2n + 1) (d) \(\frac{n}{2}\)(n3 + 4n2 - 2n + 1)
Answer» (a) \(\frac{n}{2}\)(n³+ 4n² + 4n - 1) nth term (Tn) of the given series = (2n – 1) (n + 1)² = (2n – 1) (n² + 2n + 1) = 2n³ – n² + 4n² – 2n + 2n – 1 = 2n³ + 3n² – 1 ∴ S_n = \(2\displaystyle\sum_{k=1}^{n} k^3\) + \(3\displaystyle\sum_{k=1}^{n} k^2-n\) = 2 x \(\frac{n^2(n+1)^2}{4}\) + 3 x \(\frac{1}{6}\) x n (n + 1)(2n + 1) - n = \(\frac{1}{2}\)[n² (n² + 2n +1) + (n² + n)(2n + 1) -2n] = \(\frac{1}{2}\)[n⁴ + 2n³ + n³ + 2n³ + 2n² + n² + n - 2n] = \(\frac{1}{2}\)[n⁴ + 4n³ + 4n² - n] = \(\frac{n}{2}\)(n³+ 4n² + 4n - 1)

Find the sum of the series 1.22 + 3.32 + 5.42 + .... to n terms(a) \(\frac{n}{2}\)(n3 + 4n2 + 4n - 1)(b)\(\frac{n}{2}\)(n + 1)2 (2n + 1) (c) \(\frac{n^2}{3}\)(n + 1)2 (2n + 1) (d) \(\frac{n}{2}\)(n3 + 4n2 - 2n + 1)

Answer»

(a) \(\frac{n}{2}\)(n³+ 4n² + 4n - 1)

nth term (Tn) of the given series

= (2n – 1) (n + 1)² = (2n – 1) (n² + 2n + 1)

= 2n³ – n² + 4n² – 2n + 2n – 1 = 2n³ + 3n² – 1

∴ S_n = \(2\displaystyle\sum_{k=1}^{n} k^3\) + \(3\displaystyle\sum_{k=1}^{n} k^2-n\)

= 2 x \(\frac{n^2(n+1)^2}{4}\) + 3 x \(\frac{1}{6}\) x n (n + 1)(2n + 1) - n

= \(\frac{1}{2}\)[n² (n² + 2n +1) + (n² + n)(2n + 1) -2n]

= \(\frac{1}{2}\)[n⁴ + 2n³ + n³ + 2n³ + 2n² + n² + n - 2n]

= \(\frac{1}{2}\)[n⁴ + 4n³ + 4n² - n] = \(\frac{n}{2}\)(n³+ 4n² + 4n - 1)