大师作文网(1/2)谢谢解答下列问题祝中秋快乐.设f(n)=(1/n+1)+(1/n+2)+(1/n+3)+…+(1/2n)(n
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/11 20:17:43
(1/2)谢谢解答下列问题祝中秋快乐.设f(n)=(1/n+1)+(1/n+2)+(1/n+3)+…+(1/2n)(n
f(n)有n项,则
f(n+1)-f(n)
=[(1/(n+2))+(1/(n+3))+···+(1/(2n))+(1/(2n+1))+(1/(2n+2))]
-[(1/(n+1))+(1/(n+2))+···+(1/(2n))]
=[(1/(2n+1))+(1/(2n+2))]-(1/(n+1))
=(1/(2n+1))-(1/(2n+2))
=1/(2(n+1)(2n+1))
f(n+1)-f(n)
=[(1/(n+2))+(1/(n+3))+···+(1/(2n))+(1/(2n+1))+(1/(2n+2))]
-[(1/(n+1))+(1/(n+2))+···+(1/(2n))]
=[(1/(2n+1))+(1/(2n+2))]-(1/(n+1))
=(1/(2n+1))-(1/(2n+2))
=1/(2(n+1)(2n+1))
(1/2)谢谢解答下列问题祝中秋快乐.设f(n)=(1/n+1)+(1/n+2)+(1/n+3)+…+(1/2n)(n
证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
2^n/n*(n+1)
设f(n)=1+1/2+1/3+…+1/2n 则f(n+1)-f(n)=?
设f(n)＝1n+1+1n+2+1n+3+…+13n(n∈N*)，则f（n+1）-f（n）=（ ）
f(n)=1/(n+1)+1/(n+2)+1/(n+3)……+1/2n (n∈N*),f(n+1
设f(x)=2^x/(2^x+根号2),求f(1/n)+f(2/n)+f(3/n)+.+f(n/n)(n为自然数)
设n∈N,n>1.求证：logn (n+1)>log(n+1) (n+2)
已知递推公式f(n)=(n-1)(n-2)[f(n-2)+f(n-3)+(n-3)*f(n-4)] (n>4)求通项公式
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
(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
数列a(n)=n (n+1)(n+2)(n+3), 求S(n)怎么用高中数列原理解答?