Sum of n terms of the series 13 + 33 + 53 + 73 + ..... is (a) n2 (2n2

1.	Sum of n terms of the series 13 + 33 + 53 + 73 + ..... is (a) n2 (2n2 – 1) (b) 2n2 + 3n2 (c) n3 (n – 1) (d) n3 + 8n + 4
Answer» (a) n² (2n² – 1). Let Sn = 1³ + 3³ + 5³ + 7³ + ..... upto n terms nth term of the series = (2n – 1)³ ∴ S_n = \(\displaystyle\sum_{k=1}^{n}(2k-1)^3\) = \(\displaystyle\sum_{k=1}^{n}[8k^3-1-12k^2+6k]\) = \(8\displaystyle\sum_{k=1}^{n}k^3\) - \(12\displaystyle\sum_{k=1}^{n}k^2\) + \(6\displaystyle\sum_{k=1}^{n}k-n\) = 8 \(\frac{n^2(n+1)^2}{4}\)- 12 x \(\frac{n(n+1)(2n+1)}{6}\) + 6 x \(\frac{n(n+1)}{2}\) - n = 2n² (n² + 2n + 1) – 2n (2n² + 3n + 1) + 3n(n + 1) – n = 2n⁴ + 4n³ + 2n² – 4n³ – 6n² – 2n + 3n² + 3n – n = 2n⁴ + n²= n² (2n² – 1).

Sum of n terms of the series 13 + 33 + 53 + 73 + ..... is (a) n2 (2n2 – 1) (b) 2n2 + 3n2 (c) n3 (n – 1) (d) n3 + 8n + 4

Answer»

(a) n² (2n² – 1).

Let Sn = 1³ + 3³ + 5³ + 7³ + ..... upto n terms

nth term of the series = (2n – 1)³

∴ S_n = \(\displaystyle\sum_{k=1}^{n}(2k-1)^3\) = \(\displaystyle\sum_{k=1}^{n}[8k^3-1-12k^2+6k]\)

= \(8\displaystyle\sum_{k=1}^{n}k^3\) - \(12\displaystyle\sum_{k=1}^{n}k^2\) + \(6\displaystyle\sum_{k=1}^{n}k-n\)

= 8 \(\frac{n^2(n+1)^2}{4}\)- 12 x \(\frac{n(n+1)(2n+1)}{6}\) + 6 x \(\frac{n(n+1)}{2}\) - n

= 2n² (n² + 2n + 1) – 2n (2n² + 3n + 1) + 3n(n + 1) – n

= 2n⁴ + 4n³ + 2n² – 4n³ – 6n² – 2n + 3n² + 3n – n

= 2n⁴ + n²= n² (2n² – 1).