$\frac{\frac{1}{2}.\frac{2}{2}}{1^{3}} + \frac{\frac{2}{2}.\frac{3}{2}}{1^{3} + 2^{3}} + \frac{\frac{3}{2}.\frac{4}{2}}{1^{3} + 2^{3} + 3^{3}} + ... + n$ terms equals

Question

Dr. Abubacker · Accepted Answer

(c)
$T_{n} = \frac{\frac{n(n + 1)}{2.2}}{1^{3} + 2^{3} + 3^{3} + ... + n^{3}}$
$= \frac{\frac{n(n + 1)}{4}}{\left( \frac{n(n + 1)}{2} ight)^{2}} = \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$
$	herefore T_{n} = \Sigma\left( \frac{1}{n} - \frac{1}{n + 1} ight)$
$= \left( 1 - \frac{1}{2} ight) + \left( \frac{1}{2} - \frac{1}{3} ight) + \left( \frac{1}{3} - \frac{1}{4} ight) + ... + \left( \frac{1}{n} - \frac{1}{n + 1} ight)$
$= 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

Q. $\frac{\frac{1}{2}.\frac{2}{2}}{1^{3}} + \frac{\frac{2}{2}.\frac{3}{2}}{1^{3} + 2^{3}} + \frac{\frac{3}{2}.\frac{4}{2}}{1^{3} + 2^{3} + 3^{3}} + ... + n$ terms equals

Explanation

Reviewed by Dr. Abubacker

Want more practice?