推 jknm0510a :感謝，原來是用數學歸納法... 01/16 19:20
※ 引述《jknm0510a (Kang)》之銘言：
: 想請問一下各位
: 1. Prove that 1^2-2^2+3^2-.....+(-1)^n-1*n^2= (-1)^n-1*n(n+1)/2 by induction
: for n>=1.
n=1顯然成立
假設n=k,1^2-2^2+3^2-.....+(-1)^k-1*k^2= (-1)^k-1*k(k+1)/2成立
n=k+1,
1^2-2^2+3^2-.....+(-1)^k-1*k^2 + (-1)^k*(k+1)^2
= (-1)^k-1*k(k+1)/2 + (-1)^k*(k+1)^2
= (-1)^k-1[k(k+1)/2-(k+1)^2]
= (-1)^k-1(k+1)[k/2-(k+1)]
= (-1)^k-1(k+1)[-(k+2)/2]
= (-1)^k*(k+1)(k+2)/2
得證
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