Question

（复杂）有理数指数幂计算有理数指数幂计算：(1+2^1\32)(1+2^1\16)(1+2^1\8)(1+2^1\4)(

Answer 1

(1+2^1/32)(1+2^1/16)(1+2^1/8)(1+2^1/4)(1+2^1/2)

上下同乘以(1-2^1/32),反复用平方差

=(1-2^1/32)(1+2^1/32)(1+2^1/16)(1+2^1/8)(1+2^1/4)(1+2^1/2)/(1-2^1/32)

=[1^2-(2^1/32)^2](1+2^1/32)(1+2^1/16)(1+2^1/8)(1+2^1/4)(1+2^1/2)/(1-2^1/32)

=(1-2^1/16)(1+2^1/16)(1+2^1/8)(1+2^1/4)(1+2^1/2)/(1-2^1/32)

=(1-2^1/8)(1+2^1/8)(1+2^1/4)(1+2^1/2)/(1-2^1/32)

=(1-2^1/4)(1+2^1/4)(1+2^1/2)/(1-2^1/32)

=(1-2^1/2)(1+2^1/2)/(1-2^1/32)

=(1-2)/(1-2^1/32)

=-1/(1-2^1/32)

=1/(2^1/32-1)