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Mathematics Question on Sequence and series

23−131×7+43−33+22−132×11+63−53+43−33+23−133×15+⋯+303−293+283−273+⋯+23−1315×63\begin{array}{l} \frac{2^3-1^3}{1\times7}+\frac{4^3-3^3+2^2-1^3}{2\times 11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times 15}+\cdots+\frac{30^3-29^3+28^3-27^3+\cdots+2^3-1^3}{15\times63}\end{array}

is equal to _______.

Answer

Tn=∑k=1n[(2k)3−(2k−1)3]n(4n+3)\begin{array}{l} T_n=\frac{\displaystyle\sum\limits_{k=1}^{n}\left[\left(2k\right)^3-\left(2k-1\right)^3\right]}{n\left(4n+3\right)} \end{array}
=∑k=1n4k2+(2k−1)2+2k(2k−1)n(4n+3)\begin{array}{l} =\frac{\displaystyle\sum\limits_{k=1}^{n}4k^2+\left(2k-1\right)^2+2k\left(2k-1\right)}{n\left(4n+3\right)}\end{array}
=∑k=1n(12k2−6k+1)n(4n+3)\begin{array}{l} =\frac{\displaystyle\sum\limits_{k=1}^{n}\left(12k^2-6k+1\right)}{n\left(4n+3\right)}\end{array}
=2n(2n2+3n+1)−3n2−3n+nn(4n+3)\begin{array}{l} =\frac{2n\left(2n^2+3n+1\right)-3n^2-3n+n}{n\left(4n+3\right)} \end{array}
=n2(4n+3)n(4n+3)=n\begin{array}{l} =\frac{n^2\left(4n+3\right)}{n\left(4n+3\right)}=n\end{array}
∴ Tn=n\begin{array}{l} \therefore\ T_n=n \end{array}
Sn=∑n=115Tn=15×162=120\begin{array}{l} S_n=\displaystyle\sum\limits_{n=1}^{15}T_n=\frac{15\times16}{2}=120\end{array}