1/1×4+1/4×7...1/(3n-2)×(3n+1) 1/1×3+1/3×5...1/(2n-1)×(2n+1)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/09/18 21:23:21
1/1×4+1/4×7...1/(3n-2)×(3n+1) 1/1×3+1/3×5...1/(2n-1)×(2n+1)
Sn1=1/1×4+1/4×7...1/(3n-2)×(3n+1)
Sn2=1/1×3+1/3×5...1/(2n-1)×(2n+1)
Sn1=1/1×4+1/4×7...1/(3n-2)×(3n+1)
Sn2=1/1×3+1/3×5...1/(2n-1)×(2n+1)
1/[(3n-2)×(3n+1)]
=1/3*[1/(3n-2)-1/(3n+1)]
则Sn1=1/(1×4)+1/(4×7)...1/[(3n-2)×(3n+1)]
=1/3*[1/1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]
=1/3*[1-1/(3n+1)]
=n/(3n+1)
1/[(2n-1)×(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
所以
Sn2=1/(1×3)+1/(3×5)...1/[(2n-1)×(2n+1)]
=1/2*[1/1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=n/(2n+1)
=1/3*[1/(3n-2)-1/(3n+1)]
则Sn1=1/(1×4)+1/(4×7)...1/[(3n-2)×(3n+1)]
=1/3*[1/1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]
=1/3*[1-1/(3n+1)]
=n/(3n+1)
1/[(2n-1)×(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
所以
Sn2=1/(1×3)+1/(3×5)...1/[(2n-1)×(2n+1)]
=1/2*[1/1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=n/(2n+1)
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
化简:1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
化简(n+1)(n+2)(n+3)
一道数列求和题1/2n+3/4n+5/8n+...+(2n-1)/n*2^n
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】
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 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大
1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3