Question

裂项法数列求和的一些题目，高手进

这个..

Answer 1

n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
1*2+2*3+...+(n-1)n+n(n+1)=(1/3)[1*2*3-0*1*2 + 2*3*4-1*2*3 +...+(n-1)n(n+1)-(n-2)(n-1)n +n(n+1)(n+2)-(n-1)n(n+1)] = (1/3)[n(n+1)(n+2)]=n(n+1)(n+2)/3
1*2+2*3+...+50*51=50*51*52/3=50*17*2*26=44200

n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
1*2*3+2*3*4+...+(n-1)n(n+1)+n(n+1)(n+2)=(1/4)[1*2*3*4-0*1*2*3 + 2*3*4*5-1*2*3*4 +...+(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1) + n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=(1/4)[n(n+1)(n+2)(n+3)]
=n(n+1)(n+2)(n+3)/4
------
1+2+...+n=n(n+1)/2
1/[1+2+...+n]=2/[n(n+1)] = 2/n - 2/(n+1)
1/[1+2] + 1/[1+2+3] + ... + 1/[1+2+...+(n-1)] + 1/[1+2+...+n]
=2/2 - 2/3 + 2/3 - 2/4 + ... +2/(n-1) - 2/(n) + 2/n - 2/(n+1)
=1 - 2/(n+1)
=(n-1)/(n+1)

Answer 2

先把分母换成等差数列通项式的写法，然后裂项化简

Answer 3

1+2=（1+2）*2/2=1/2*（2*3）
1+2+3=（1+3）*3/2=1/2*（3*4）
。。
1+2+3+..+n=(1+n)*n/2(等差数列求和)=1/2*（n*（n+1））
原式可化为：2{1/(2*3）+ 1/(3*4) + ..... + 1/(n*(n+1))}
=>2{1/2 - 1/3 + 1/3 - 1/4 +...+ 1/n - 1/(n+1)}
=>2*{1/2 - 1/(n+1)}

Answer 4

2【1/（1+2）+1/(2+3)+1/(3+4)+1/(1+2+3+...+n）】-1
=2[（1-1/2+1/2-1/3+1/3+。。。+1/n-1/(1+n)]-1
=2[1-1/(1+n)]-1
=1-2/(1+n)