\[\frac{2^3-1^3}{1\times7}+\frac{4^3-3^3+2^3-1^3}{2\times11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times15}+\cdots+\frac{30^3-29^3+\cdots+2^3-1^3}{15\times63}\]
is equal to \_\_\_\_\_\_\_\_.
Answer: 120
Solutions · 1
Official solution
The nth numerator is k=1∑n[(2k)3−(2k−1)3]=k=1∑n(12k2−6k+1)=n(4n2+3n)⋅⋯=n(2n+1)(2n−1)+⋯; combined with the denominator n(4n+3), each term simplifies exactly to Tn=n.
\[S = \sum_{n=1}^{15}n = \frac{15\times16}{2} = 120.\]