Question

（1-1/2^2）(1-1/3^2) (1-1/4^2)…（1-1/10^2）要详细过程，好让我明白，最好每个步骤都有解释。

Answer 1

(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/10²)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)(1+1/4)(1-1/4)...(1+1/10)(1-1/10) 每个括号内运用平方差公式
=(3/2)(1/2)(4/3)(2/3)(5/4)(3/4)...(11/10)(9/10)
=[(3/2)(4/3)(5/4)...(11/10)][(1/2)(2/3)...(9/10)]
=[(3×4×5×...×11)/(2×3×4×...×10)]×[(1×2×...×9)/(2×3×...×10)]
=(11/2)(1/10)
=11/20

Answer 2

你好：

（1-1/2^2）(1-1/3^2) (1-1/4^2)…（1-1/10^2）
=（1+1/2）(1-1/2)(1+1/3)(1-1/3)(1+1/4)(1-1/4)......(1+1/10)(1-1/10)
=1/2*3/2*2/3*4/3*3/4*5/4......*9/10*11/10 相邻分数分子分母互相约掉
=1/2*11/10
=11/20

希望对你有帮助！

Answer 3

A1 = 1 - 1/[(1+1)^2] = [(1+1)^2 - 1]/[(1+1)^2] = [(1+1+1)(1+1-1)]/[(1+1)^2] = (1+2)*1/[(1+1)^2]

A2 = (1+3)(1+1)/[(1+2)^2]

A3 = (1+4)(1+2)/[(1+3)^2]

A1 * A2 = (1+2)*1/[(1+1)^2] * (1+3)(1+1)/[(1+2)^2] = [(1+3)*1] / [(1+1)(1+2)] = [(2+2)*1] / [(2+1)(1+1)]

A1*A2*A3 = [(1+3)*1] / [(1+1)(1+2)] * (1+4)(1+2)/[(1+3)^2] = [(1+2+2)*1] / [(1+2+1)(1+1)] = [(3+2)*1] / [(3+1)(1+1)]

A1*A2*A3*……*An = (n+2)/[2(n+1)]

A1*A2*……*A9 = [(9+2)*1] / [(9+1)(1+1)] = 11/20

Answer 4

解：
（1-1/2^2）(1-1/3^2)(1-1/4^2)……(1-1/10^2)
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)……(1-1/10)(1+1/10)
=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)……(9/10)(11/10)
=(1/2)*(11/10) （中间的都约分约掉了）
=11/20

Answer 5

很高兴为您解答：
（1-1/2^2）(1-1/3^2)(1-1/4^2)...(1-1/10^2)
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/10)(1+1/10)
=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)...(9/10)(11/10)
=(1/2)(11/10)
=11/20
谢谢，祝你开心
有帮助记得采纳哦