求和 1⼀2!+2⼀3!+3⼀4!+....+n⼀(n+1)!

2024-11-26 13:27:32
推荐回答(2个)
回答1:

首先把n(n+1)拆成n^2+n,然后每一项都以此类推
原式=(1^2+1)+(2^2+2)+(3^2+3)……(n^2+n)
=(1^2 + 2^2 + 3^2 + 4^2 + …… + n^2)+(1+2+3+4+……+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3

回答2:

n/(n+1)!=(n+1-1)/(n+1)!=(n+1)/(n+1)!-1/(n+1)!
=1/n!-1/(n+1)!
1/2!+2/3!+3/4!+....+n/(n+1)!
=(1/1!-1/2!)+(1/2!-1/3!)+(1/3!-1/4!)+......+!)+......+[1/(n-1)!-1/n!]+[1/n!-1/(n+1)!]
=1-1/(n+1)!