Question

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 1

(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).