1*3*5+5*7*9+……97*99*101
问题描述：
1*3*5+5*7*9+……97*99*101
答
待求和式=∑(4n+1)(4n+3)(4n+5),n从0到24.用裂项相消法,(4n+1)(4n+3)(4n+5) = 1/8 *[(4n+1)(4n+3)(4n+5)(4n+7) - (4n-1)(4n+1)(4n+3)(4n+5)]所以∑(4n+1)(4n+3)(4n+5),n从0到24=1/8 * [97*99*101*103 - (-1)*1*3*5]=...