1/1+2+1/2+3+1/3+4……+1/n(n

1/1+2+1/2+3+1/3+4……+1/n(n+1)=？

1个回答

n阶行列式计算|1 2 3.n-1 n||2 1 3.n-1 n||2 3 1.n-1 n||.||2 3 4. 1 n
1\n(n+1)+1\(n+1)(n+2)+1\(n+2)(n+3)+1\(n+3)(n+4)
用数学归纳法证明：1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.+1/[n(n+)(n+)]
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3
定义n!=1*2*3*.*(n-1)*n,例如4!=1*2*3*4,化简1/2!+2/3!+n/(n+1)!
4/(1*2*3)+7/(2*3*4)+.+(3n+1)/[n(n+1)(n+2)]
1/2-1/n+1＜1/2^2+1/3^2+……+1/n^2＜n-1/n(n=2,3,4,5,6.
计算:1/(1*2*3)+1/(2*3*4)+.+1/[n(n+1)(n+2)]