數學歸納法

When n=1,

L.H.S.=(-1^1)[2(1)-1]=-1

R.H.S=(-1^1)(1)=-1

Therefore n=1 is ture.

Assume n=k is ture, i.e. -1+3-5+...+(-1)^k(2k-1)=(-1)^k(k)

When n=k+1

L.H.S = -1+3-5+...+(-1)^k(2k-1)+(-1)^(k+1)(k+1)

= (-1)^k(k)+(-1)^(k+1)(k+1)

= R.H.S

Therefore n=k+1 is ture for all positive natural numbers n

By the principle of Mathemactical Induction,S(n) is ture for all positive natural numbers n.

設P(n)為命題,'-1+3-5+...+(-1)^n(2n-1)=(-1)^n(n)'
當n=1時,
左方=(-1)^1[2(1)-1]=-1
右方=-1^1(1)=1
所以P(1)成立
設P(k)成立,即'-1+3-5+...+(-1)^k(2k-1)=(-1)^k(k)'
當n=k+1時,
左方=-1+3-5+...+(-1)^k(2k-1)+(-1)^(k+1)[2(k+1)-1]
=(-1)^k(k)+(-1)^(k+1)[2(k+1)-1]
=(-1)^(k)(k)+[(-1)^k(-1)](2k+1)
=(-1)^(k)[k-(2k+1)]
=(-1)^(k)(-k-1)
=(-1)^(k)(-1)(k+1)
=(-1)^(k+1)(k+1)
=右方
所以P(k+1)成立
根據數學歸納法,對任意正整數n,P(n)都成立