n(n-1)/2

2^n/n*(n+1)2^n/n*(n+1)2^n/n*(n+1)1/2*f(1/2)=(1/2)^2+3*(1/2)^3...+(2n-1)*(1/2)^(n+1)f(1/2)-1/2*f(1/2)=1/2+2*(1/2)^2+2*(1/

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n先证明

lim[n/(n*n+1*1)+n/(n*n+2*2)+...+n/(n*n+n*n)],当x趋向无穷大时,怎么求极限,lim[n/(n*n+1*1)+n/(n*n+2*2)+...+n/(n*n+n*n)],当x趋向无穷大时,怎么求极限,

(n+2)!/(n+1)!(n+2)!/(n+1)!(n+2)!/(n+1)!(n+2)!/(n+1)!=[1×2×3×……×(n+2)]/[1×2×3×……×(n+1)]=n+2二恶英提供

1+(n+1)+n*(n+1)+n*n+(n+1)+1=2n^2+3n+3为什么等于2n^2+3n+3怎么推算的很多数学的都忘了,还有设计道德只是范涛是那个年级的什么篇章1+(n+1)+n*(n+1)+n*n+(n+1)+1=2n^2+3n

[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[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简原式=(3n²+3n+2n

[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[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简分子是6还是6+n(n+2)

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化简n分之n-1+n分之n-2+n分之n-3+.+n分之1(n-1)/n=1-1/n各项都可以这么化那么S=(n-1)/n+

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化简n分之n-1+n分之n-2+n分之n-3+.+n分之1(1+2+3+.+n-1)/n=(1+n-1)*(n-1)/2/n

f(x)=e^x-x求证(1/n)^n+(2/n)^n+...+(n/n)^nf(x)=e^x-x求证(1/n)^n+(2/n)^n+...+(n/n)^nf(x)=e^x-x求证(1/n)^n+(2/n)^n+...+(n/n)^n令ak

排列组合C(0n)+C(1n)+C(2n)+...+C(n-1n)+C(nn)（n∈N*）的值,并证明你的结果.括号内前一个数在C上方,后一个数在C下方...值为2^n,我想知道怎么证明.排列组合C(0n)+C(1n)+C(2n)+...+

化简(n+1)(n+2)(n+3)化简(n+1)(n+2)(n+3)化简(n+1)(n+2)(n+3)设n+2=x所以(n+1)(n+2)(n+3)=(x-1)*x*(x+1)=(x^2-1)*x=x^3-x将n+2=x代入,得n^3+3n

n*【n+1】*【n+2】化简成什么?n*【n+1】*【n+2】化简成什么?n*【n+1】*【n+2】化简成什么?n(n+1)(n+2)=(n²+n)(n+2)=n³+2n²+n²+2n=n³

2n/（n+1）n!2n/（n+1）n!2n/（n+1）n!2n/(n+1)!

n(n+1)(n+2)等于多少?n(n+1)(n+2)等于多少?n(n+1)(n+2)等于多少?n（n+1）（n+2）=（n平方+n）（n+2）=n^3+3n^2+2n

n+(n-1)÷2×n求化简n+(n-1)÷2×n求化简n+(n-1)÷2×n求化简n+n(n-1)/2=n+1/2(n^2-n)=1/2(2n+n^2-n)=1/2(n^2+n)

n(n+1)(n+2)什么意思n(n+1)(n+2)什么意思n(n+1)(n+2)什么意思 

1^n+2^n+...+2013^n1^n+2^n+...+2013^n1^n+2^n+...+2013^n因为1^n

2n*1/2^n=n/2^(n-1)?2n*1/2^n=n/2^(n-1)?2n*1/2^n=n/2^(n-1)?成立

化简[n^2+(n+1)^2]/n(n+1)化简额化简[n^2+(n+1)^2]/n(n+1)化简额化简[n^2+(n+1)^2]/n(n+1)化简额[n^2+(n+1)^2]/n(n+1)=n/(n+1)+(n+1)/n