1/6+1/12+…+1/n(n+1)=2003/2004,求n
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(n+1)^n-(n-1)^n=?
推导 n*n!=(n+1)!-n!
Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗?
用数学归纳法证明 6+2*9+3*12+…+n(3n+3)=n(n+1)(n+2)
用数学归纳法证明:1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2)
1*N+2*(N-1)+3*(N-2)+...+N*1=1/6N(N+1)(N+2)
e^(1/n)+e^(2/n)+e^(3/n)+…+e^(n-1/n)+e^(n/n)=?
[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
9题 = 101 (n+1)!- = n*n!n(n+1)!- n*n!
f(x)=e^x-x 求证(1/n)^n+(2/n)^n+...+(n/n)^n
n^(n+1/n)/(n+1/n)^n
证明:(n+1)n!= (n+1)!
证明:1+2+3+……+n=1/6n(n+1)(2n+1)
为什么 [ln(n)]'/n'=1/n
1+2+3+4+……+n=n(n+1)(2n+1)/6
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
2^n/n*(n+1)