(1-1/2^2)(1-1/3^2)...(1-1/n^2)
=(2+1)(2-1)(3+1)(3-1)(4+1)(4-1)…(n+1)(n-1)/(n!)^2
=[3*4*5*…(n+1)]*[1*2*3…(n-1)]/(n!)^2
=(n+1)!(n-1)!/[2*(n!)^2]
=(n+1)*n!*n!/[2*n*(n!)^2
=(n+1)/(2n)
求证(1-1/2^2)(1-1/3^2)...(1-1/n^2)=(n+1)/2n,
4个回答
相关问题
-
归纳法求证1^2/(1*3)+2^2/(3*5)+...+n^2/(2n-1)(2n+1)=n(n+1)/[2(2n+1
-
求证：1+1/2+1/3+···+1/2^n＞n+2/2 (n≥2）
-
求证1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n
-
求证:1-1/2+1/3-1/4+...-1/2n=1/(n+1)+1/(n+2)+...+1/2n
-
求证(n+1)(n+2)(n+3)……(n+n)=2^n*1*3*……*(2n-1)
-
设f(n)=1+1/2+1/3+…+1/n(n∈N*,n>1)求证2f(2^n)>n+2
-
求证1!+2*2!+3*3!+…+n*n!=(n+1)!-1
-
f(n)=1+1/2+1/3+1/4...+1/n,求证f(2∧n)>n/2,n∈N+
-
求证：1*2+2*5+3*8+…+n(3n-1)=n^2(n+1)
-
求证n≥2 ,n为正整数时,求证4/7≤1/（n+1)+1/(n+2)+...+1/(2n-1)+1/2n＜√2/2