1/2+3/2^2+5/2^3+...+(2n-1)/2^n=
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1/2+3/2^2+5/2^3+...+(2n-1)/2^n=
![1/2+3/2^2+5/2^3+...+(2n-1)/2^n=](/uploads/image/z/15684944-32-4.jpg?t=1%2F2%2B3%2F2%5E2%2B5%2F2%5E3%2B...%2B%282n-1%29%2F2%5En%3D)
t=1/2+3/2^2+5/2^3+...+(2n-3)/2^(n-1)+(2n-1)/2^n
2t=1+3/2+5/2^2+...+(2n-3)/2^(n-2)+(2n-1)/2^(n-1)
相减有
t=1+[2/2+2/2^2+2/2^3+...2/2^(n-1)]-(2n-1)/2^n
=1+[(1/2)^(n-1)-1]/(1/2-1)]-(2n-1)/2^n
=1+2-2^(2-n)-(2n-1)/2^n
=3-(2n+3)/2^n
2t=1+3/2+5/2^2+...+(2n-3)/2^(n-2)+(2n-1)/2^(n-1)
相减有
t=1+[2/2+2/2^2+2/2^3+...2/2^(n-1)]-(2n-1)/2^n
=1+[(1/2)^(n-1)-1]/(1/2-1)]-(2n-1)/2^n
=1+2-2^(2-n)-(2n-1)/2^n
=3-(2n+3)/2^n
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
证明(1+2/n)^n>5-2/n(n属于N+,n>=3)
1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
化简(n+1)(n+2)(n+3)
2^n/n*(n+1)
证明:1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+)
计算:n(n+1)(n+2)(n+3)+1
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
(1/(n^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大
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)..
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?