裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题 裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/15 21:47:01
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题裂项相消法求和(1)an=1/(2n+1)(2n
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题 裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题 裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
(1)an=(1/2)[1/(2n+1)-1/(2n+3)],
∴a1+a2+……+an
=(1/2)[1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)]
=(1/2)[1/3-1/(2n+3)]
=n/(6n+9).
先把an裂项,求和时注意消去哪些项,留下哪些项.余者类推.
剩下的题目留给您练习.
电视机厂8月份共生产电视机2400台,其中彩电1500台,其余的是黑白电视机.这个厂8月份生产的黑白电视机占电视机总数的几分之几?
裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题 裂项相消法求和(1)an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)(4)an=2/n(n+1)四道题
裂项相消法求和所有公式1/(2n+1)(2n+3)(2n+5)
求和1/1×2+1/2×3+.+1/n(n+1)用裂项相消法如题,
已知等差数列an前n项和为Sn,Sn=n^2,求和1/(a1a2)+1/(a2a3)+.+1/[(an-1an] (n≥2 )老师说,用裂项相消法,求完整过程,
高二数列求和错位相消法解题求an=(2n+1)^5(2n-1)的前n项和怎么用错位相消法来做啊?应该是这个:an=(2n+1)·5^(2n-1),求它的前n项和
数列求和中拆项相消法怎么拆,有什么公式吗,比如:1/(2n-1)(2n+1)
An=1/n^2 数列求和An=1/n^2 数列(An)求和
裂项相消法常见公式1/n(n+1)(n+2)=?
用裂项相消法做下列3个题 1.an=1/n(n+1) 2.an=1/(3n+1)(3n+2) 3.an=1/n²+2n
高二数列求和 An=(2n+1)^2/[2n(n+1)] 数列求和
an=(2n+1)(1/2)^n-1求和
an=(2n)(1/2)^n-1求和
请问数列an=n/(2n+1)如何求和
数列an=(n(n+1))/2 求和
裂项相消法 隔项相消原式=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+.+(1/n-1/(n+2))]请问怎样求和?请写出具体计算过程 .1/2(3/2-1/(n+1)-1/(n+2)) 请问为什么是1/(n+1)?
求和sn=1/2+2/2的平方+3/2的3次方+…+n-1/2的(n-1)次方+n/2的n次方属于用错位相消法求一类数列前n项的和
裂项相消法求和例题
数列求和裂项相消法