4/3×6/5×8/7×······×(2n+2)/(2n+1)求积

liulongjiang2022-10-04 11:39:542条回答

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午后阳光2003 共回答了13个问题 | 采纳率92.3%: 2^(2n+1)[(n+1)!]^2/(2n+2)!; 1年前

勇气在哪里共回答了3个问题 | 采纳率: 原式
=(4*6*...*(2n+2))/(3*5*...*(2n+1))
=(4*6*...*(2n+2))^2/(3*4*5*...*(2n+2))
=2^2n*(2*3*...*(n+1))^2/(3*4*5*...*(2n+2))
=2^2n*((n+1)!)^2/((2n+2)!/2)
=2^(2n+1)*((n+1)!)^2/(2n+2)!; 1年前

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