1/2(2-1)+1/3(3-1)+1/4(4-1)+……+1/n(n-1)=?
4个回答
写出通项公式:a=1/n (n-1) 用拆项法拆成:a=(1/n-1)-(1/n) n≥2
最后用叠加法求和得:S=1-1/n
相关问题
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.+1/[n(n+)(n+)]
设f(n)=1-(1/(2*2))+(1/(3*3))-(1/(4*4))…… +(-1)^(n-1) * (1/(n*
已知:1/1*2=1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4,…,1/n*(n+1)=1/n-1
1/1*2=1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4,.1/n(n+1)=1/n+1,以上等式
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\n(n+1)+1\(n+1)(n+2)+1\(n+2)(n+3)+1\(n+3)(n+4)
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)]
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
1/1+2+1/2+3+1/3+4……+1/n(n+1)=?