1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 13:18:30
1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)
当n=1时显然成立
设当n=k时,有1*k+2(k-1)+3(k-2)+······+k*1=1/6*k(k+1)(k+2)
当n=k+1时,有
1*(k+1)+2(k+1-1)+3(k+1-2)+······+(k+1)*1
=1*k+2(k-1)+3(k-2)+······+k*1 + (1+2+3+……+k+(k+1))
=1/6*k(k+1)(k+2)+(k+1)(k+2)/2
=1/6*(k+1)(k+2)(k+3)
即n=k+1时成立
故有……
一、n=1时,1=1成立
二、令n=k时上式成立,则n=k+1时
1*(n+1)+2*n+3(n-1)+······+n*2+(n+1)*1
=1*n+1+2(n-1)+2+3(n-2)+3+······+n*1+n+(n+1)*1
=1*n+2(n-1)+3(n-2)+······+n*1+1+2+3+······+(n+1)
=1/6*n(n+1)(n+...
全部展开
一、n=1时,1=1成立
二、令n=k时上式成立,则n=k+1时
1*(n+1)+2*n+3(n-1)+······+n*2+(n+1)*1
=1*n+1+2(n-1)+2+3(n-2)+3+······+n*1+n+(n+1)*1
=1*n+2(n-1)+3(n-2)+······+n*1+1+2+3+······+(n+1)
=1/6*n(n+1)(n+2)+(n+1)(n+2)/2
=1/6*(n+3)(n+1)(n+2)
由一、二得,等式成立
收起
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
2^n/n*(n+1)
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
化简n分之n-1+n分之n-2+n分之n-3+.+n分之1
化简n分之n-1+n分之n-2+n分之n-3+.+n分之1
lim[n/(n^2+1^2)+n/(n2+2^2)+···n/(n^2+n^2)] n->无穷大
化简(n+1)(n+2)(n+3)
当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)...1=n
limn趋近于无穷2·5^n+3^n/5^n+1+2^n+1
求级数1/[3^n+(-2)^ n]·x^n/n的收敛域
1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)
lim2^n +3^n/2^n+1+3^n+1
3(n-1)(n+3)-2(n-5)(n-2)
n(n+1)(n+2)(n+3)+1 因式分解
n(n+1)(n+2)(n+3)+1等于多少
lim(n+3)(4-n)/(n-1)(3-2n)
lim(n^3+n)/(n^4-3n^2+1)