AMATHS的mi題目啊..急

因為我用中文讀，我打中文，你轉番d句子做英文吧 =)
唔好意思

設P(n)為命題
"1*2/2*3+2*2^2/3*4+...n*2^n/(n+1)(n+2)=2^n+1/n+2-1"
當n=1
左方=1*2^1/(1+1)(1+2)=2/6=1/3
右方=2^1+1/1+2-1=4/3-1=1/3

假設命題於n=k時成立
"1*2/2*3+2*2^2/3*4+...k*2^k/(k+1)(k+2)=[2^(k+1)/k+2]-1"
當n=k+1時
左方=1*2/2*3+2*2^2/3*4+...k*2^k/(k+1)(k+2)+[(k+1)*2^(k+1)]/(k+1+1)(k+1+2)
=[2^(k+1)/(k+2)]-1+[(k+1)*2^(k+1)]/(k+2)(k+3)
=[2^(k+1)(k+3)+(k+1)*2^(k+1)]/(k+2)(k+3)-1
=[2^(k+1)(k+3+k+1)]/(k+2)(k+3)-1
=[2^(k+1)*2(k+2)]/(k+2)(k+3)-1
=2^(k+1)*2/(k+3)-1
=2^(k+1+1)/(k+3)-1
=[2^(k+2)/(k+3)]-1

右方=[2^(k+1+1)/k+1+2]-1
=[2^(k+2)/(k+3)]-1

所以左方=右方
根據M.I,命題對所有正整數n都成立