The sum 1⋅12−2⋅32+3⋅52−4⋅72+5⋅92−⋯+15⋅292 is \_\_\_\_\_\_\_\_.
Answer: 6952
Solutions · 1
Official solution
Separate the odd-placed (positive) and even-placed (negative) terms:
\[S = \left(\sum_{n=1}^{8}(2n-1)(4n-3)^2\right) - \left(\sum_{n=1}^{7}(2n)(4n-1)^2\right).\]
Evaluating each sum using standard formulas gives 29856−22904=6952.