※ 引述《qwerty147852 (天橋下說書人)》之銘言：
: 想請問各位前輩
: 3.4題要如何導出
: http://imgur.com/oFpMIxT
: 感謝
(c)
check n = 0 is true
assume n = k, 1+x+x^2+ ... + x^k = [x^(k+1)-1]/(x-1) is true
n = k+1, 1+x+x^2+ ... + x^k + x^(k+1) = [x^(k+1)-1]/(x-1) + x^(k+1)
= [x^(k+1)-1]/(x-1) + x^(k+1)(x-1)/(x-1)
= [x^(k+1)-1]/(x-1) + [x^(k+2)-x^(k+1)]/(x-1)
= [x^(k+2)-1]/(x-1) is true
so by induction, .......
(d)
check n = 1 is true
assume n = k, 1*1+2*2+3*2^2+...+k*2(k-1) = (k-1)2^k + 1 is true
n = k+1, 1*1+2*2+3*2^2+...+k*2(k-1)+(k+1)*2^k
=(k-1)2^k + 1 + (k+1)*2^k
=k*2^k - 2^k + 1 + k*2^k + 2^k = 2k*2^k = k*2(k+1)
so by induction, .......
應該降吧
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 203.68.75.117
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1413797847.A.5DF.html