
利用因式分解法探究:(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/n²)的值。
1个回答
展开全部
(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/n²)
= (2^2-1)/2^2*(3^2-1)/3^2*(4^2-1)/4^2...(n^2-1)/n^2
= (2-1)(2+1)/2^2*(3-1)(3+1)/3^2*(4-1)(4+1)/4^2...(n-1)(n+1)/n^2
=(n+1)/n
= (2^2-1)/2^2*(3^2-1)/3^2*(4^2-1)/4^2...(n^2-1)/n^2
= (2-1)(2+1)/2^2*(3-1)(3+1)/3^2*(4-1)(4+1)/4^2...(n-1)(n+1)/n^2
=(n+1)/n
追问
为什么(2-1)(2+1)/2^2*(3-1)(3+1)/3^2*(4-1)(4+1)/4^2...(n-1)(n+1)/n^2
=(n+1)/n ?...
追答
(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/n²)
= (2^2-1)/2^2*(3^2-1)/3^2*(4^2-1)/4^2...(n^2-1)/n^2
= (2-1)(2+1)/2^2*(3-1)(3+1)/3^2*(4-1)(4+1)/4^2...(n-1)(n+1)/n^2
= 1*3/2^2*2*4/3^2*3*5/4^2...(n-1)(n+1)/n^2
=(n+1)/n
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