$\frac{2^{3}-1^{3}}{1 imes 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 imes 11}+\frac{6^{3}-5^{3}+4^{3}-

Question

$\frac{2^{3}-1^{3}}{1 	imes 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 	imes 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 	imes 15}+\ldots+$ $\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots .+2^{3}-1^{3}}{15 	imes 63}$ is equal to

Professor PC Thomas Classes · Answer

(120) : Then $n^{	ext {th }}$ term of the sequence can be represented as\ $T_{n}=\frac{-1^{3}+2^{3}-3^{3}+4^{3}+\ldots+(2 n)^{3}}{n(4 n+3)}$\ $=\frac{2\binom{2^{3}+4^{3}+6^{3}+}{\ldots+(2 n)^{3}}-\binom{1^{3}+2^{3}+3^{3}}{\ldots+(2 n)^{3}}}{n(4 n+3)}$\ $=\frac{2^{4}\left(\frac{n(n+1)}{2}ight)^{2}-n^{2}(2 n+1)^{2}}{n(4 n+3)}$\ $=\frac{4 n^{2}(n+1)^{2}-n^{2}(2 n+1)^{2}}{n(4 n+3)}$\ $=\frac{n^{2}\left(4 n^{2}+8 n+4-4 n^{2}-4 n-1ight)}{n(4 n+3)}$\ $=\frac{n^{2}(4 n+3)}{n(4 n+3)}=n$\ $\Rightarrow \quad \sum_{n=1}^{15} T_{n}=\sum_{n=1}^{15} n=\frac{15 	imes 16}{2}=120$\