早教吧作业答案频道 -->数学-->
用数学归纳法证明1/1*2+1/3*4+…+1/(2n-1)*2n=1/(n+1)+1/(n+2)+…+1/(n+n)
题目详情
用数学归纳法证明 1/1*2+1/3*4+…+1/(2n-1)*2n=1/(n+1)+1/(n+2)+…+1/(n+n)
▼优质解答
答案和解析
① n=1时,左=1/1*2=1/2 ,右=1/(1+1)=1/2 ∴左=右 等式成立
②
假设 n=k 时等式成立,即有 1/1*2+1/3*4+…+1/(2k-1)*2k=1/(k+1)+1/(k+2)+…+1/(k+k)
n=k+1 时 左= 1/1*2+1/3*4+…+1/(2k-1)*2k + 1/(2k+1)*(2k+2)
= 1/(k+1)+1/(k+2)+…+1/(k+k) + 1/(2k+1)*(2k+2)
= 1/(k+2)+…+1/(k+k) + [ 1/(k+1)+ 1/(2k+1)*(2k+2) ]
= 1/(k+2)+…+1/(k+k) + [ 1/(k+1) ] [ 1 + 1/2 (2k+1) ]
= 1/(k+2)+…+1/(k+k) + 1/(2k+1) + 1/(2k+2)
= 1/[ (k+1)+1] + … + 1/(k+k) + 1/[(k+1)+k] + 1/[(k+1)+(k+1)] =右
等式也成立
由①②得,对于n∈N* ,等式成立
②
假设 n=k 时等式成立,即有 1/1*2+1/3*4+…+1/(2k-1)*2k=1/(k+1)+1/(k+2)+…+1/(k+k)
n=k+1 时 左= 1/1*2+1/3*4+…+1/(2k-1)*2k + 1/(2k+1)*(2k+2)
= 1/(k+1)+1/(k+2)+…+1/(k+k) + 1/(2k+1)*(2k+2)
= 1/(k+2)+…+1/(k+k) + [ 1/(k+1)+ 1/(2k+1)*(2k+2) ]
= 1/(k+2)+…+1/(k+k) + [ 1/(k+1) ] [ 1 + 1/2 (2k+1) ]
= 1/(k+2)+…+1/(k+k) + 1/(2k+1) + 1/(2k+2)
= 1/[ (k+1)+1] + … + 1/(k+k) + 1/[(k+1)+k] + 1/[(k+1)+(k+1)] =右
等式也成立
由①②得,对于n∈N* ,等式成立
看了 用数学归纳法证明1/1*2+...的网友还看了以下:
错位求和法求和:Sn=5*2+3*2²+1*2³+...+(-2n+7)*2^n我算到了-2*(2 2020-04-05 …
高分求一道初三二次根式题!已知:实数m、n、p满足m+n+|√(p-1)-1|=4√m-2+2√n 2020-06-06 …
点A(0,2/n),B(0,-2/n),c(4+2/n,0)其中n正整数,用n表示三角形ABC面积 2020-06-06 …
数列{an}中,an=-2n+29n+3,则其中最大项为()107108865/8109an=-2 2020-07-02 …
lim(-5)^n+(3)^n+2/(-5)^n+1+3^nan=1/n(n+1)1≤n≤33×1 2020-07-16 …
如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3顺便再证明一下1x2+2x3+ 2020-07-20 …
数列极限N->∞tan(pi/4+2/N)expNtan(π/4+2/n)外面还有N次方呢 2020-07-23 …
二项式系数各项系数(5x-√x)^n的展开时的各项系数和为M,二项式系数之和为N,若M-N=240 2020-07-31 …
(1)10^7除以(10^3除以10^2)(2)(x-y)^3*(x-y)^2*(y-x)(3)4* 2020-11-01 …
对于任意自然数,定义n!=1×2×……×n,如4!=1×2×3×4.那么,+2!+3!+4!+5!= 2020-11-07 …