Question

用数学归纳法解题 (1+1)(1+1/2)…(1-1/2ⁿ)≤6(1-1/2ⁿ),n∈N*,求证

最后一步证明:n=k+1时的详细过程

Answer 1

n=k时，(1+1)(1+1/2)…(1-1/2^k)<=6(1-1/2^k),n=k+1时，(1+1)(1+1/2)…(1-1/2^k)(1-1/2^(k+1))<=6(1-1/2^k)(1-1/2^(k+1))=6[1-1/2^(k+1)-(1/2^k-1/2^(2k+1))]
(1/2^k-1/2^(2k+1))>0,故6[1-1/2^(k+1)-(1/2^k-1/2^(2k+1))]<6(1-1/2^(k+1)),即(1+1)(1+1/2)…(1-1/2^
k)(1-1/2^(k+1))<6(1-1/2^(k+1))，n=k+1时命题成立。

追问

我的问题打错了!不好意思!(1+1)(1+1/2)…(1+1/2ⁿ)≤6(1-1/2ⁿ),n∈N*,求证

追答

前面是减号，到后面怎么变加号了