1*3*5+5*7*9+……97*99*101
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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
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
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