Question

Sum to n terms of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5}$+….. is –

• A. $\frac{n}{24}$ (n² + 9n + 14)
• B. $\frac{n}{24}$ (2n² + 7n + 15)
• C. $\frac{n}{24}$ (2n² + 9n + 13)
• D. $\frac{n}{24}$ (n² + 11n + 12)

Answer 1

Answer: (C)

$\frac{n}{24}$ (2n² + 9n + 13)

Answer 2

Let t_r denote the rth term of the series, then

t_r = $\frac{1^{3} + 2^{3} + ... + r^{3}}{1 + 3 + .... + (2r - 1)}$ = $\frac{\frac{1}{4}r^{2}(r + 1)^{2}}{r^{2}}$= $\frac{1}{4}$ (r + 1)²

̃ $\sum_{r = 1}^{n}t_{r} = \frac{1}{4}\sum_{r = 1}^{n}{(r + 1)^{2}}$= $\frac { 1 } { 4 }$ $\left\lbrack \sum_{r = 1}^{n + 1}{r^{2} - 1} \right\rbrack$

= $\frac { 1 } { 4 }$ $\left\lbrack \frac{(n + 1)(n + 2)(2n + 3)}{6} - 1 \right\rbrack$

= $\frac{1}{24}$ [2n³ + 9n² + 13n + 6 – 6]

= $\frac{n}{24}$ (2n² + 9n + 13).