1*3*5+5*7*9+……97*99*101

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 07:09:37

1*3*5+5*7*9+……97*99*101
1*3*5+5*7*9+……97*99*101

1*3*5+5*7*9+……97*99*101
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]
=1/8 *[999* 10^5 + 24]
=12487503