(1/2+3/（2^2）+.(2n-1)/（2^n）)n趋向于无穷大的极限
1个回答
展开全部
设f(x) = 1 + x + x^2 + . + x^n + . = 1/(1-x)
xf(x^2) = x + x^3 + x^5 + . + x^(2n+1) + . = x/(1-x^2)
对上式派信漏求导函数得
1 + 3x^2 + 5x^4 + . + (2n+1)x^(2n) + . = (x^2+1)/(x^2-1)^2
所以
1+ 3x + 5x^2 + . + (2n+1)x^n + . = (x+1)/(x-1)^2
乘以x得
x + 3x^2 + 5x^3 + . + (2n+1)x^(n+1) + . = x(x+1)/(x-1)^2
令尘烂x = 1/坦肆2得
原式 = 1/2 * 3/2 / (1/4) = 3
xf(x^2) = x + x^3 + x^5 + . + x^(2n+1) + . = x/(1-x^2)
对上式派信漏求导函数得
1 + 3x^2 + 5x^4 + . + (2n+1)x^(2n) + . = (x^2+1)/(x^2-1)^2
所以
1+ 3x + 5x^2 + . + (2n+1)x^n + . = (x+1)/(x-1)^2
乘以x得
x + 3x^2 + 5x^3 + . + (2n+1)x^(n+1) + . = x(x+1)/(x-1)^2
令尘烂x = 1/坦肆2得
原式 = 1/2 * 3/2 / (1/4) = 3
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
