M2 Mathematical Induction

[匿名]

2012-04-10 11:27 pm

Pls. 求幫幫忙
因小弟點數已不多,一次問2題, 希望不要見怪

Prove, by mathematical induction, that for all position integers n,

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回答 (2)

tttfffkkk

2012-04-11 12:03 am

✔ 最佳答案

12.
When n=1
L.H.S.=1
R.H.S.=1(2)(3)/6=1=L.H.S.
Assume k(1)+(k-1)(2)+(k-2)(3)+...+3(k-2)+2(k-1)+1(k)=k(k+1)(k+2)/6
When n=k+1
L.H.S.=(k+1)(1)+(k)(2)+(k-1)(3)+...+3(k-1)+2(k)+1(k+1)
=k(k+1)(k+2)/6+1+2+3+...+k+1
=k(k+1)(k+2)/6+(k+1)(k+2)/2
=(k+1)(k+2)(k+3)/6=R.H.S.
13.
When n=1
L.H.S.=1(2)=2
R.H.S.=1(1)(2)=2=L.H.S.
Assume 1(2)-2(4)+3(6)-4(8)+...+[(-1)^(k-1)](k)(2k)=[(-1)^(k-1)](k)(k+1)
When n=k+1
L.H.S.=1(2)-2(4)+3(6)-4(8)+...+[(-1)^(k-1)](k)(2k)+[(-1)^k](k+1)(2k+2)
=[(-1)^(k-1)](k)(k+1)+[(-1)^k](k+1)(2k+2)
=-[(-1)^k](k)(k+1)+[(-1)^k](k+1)(2k+2)
=[(-1)^k](k+1)[-k+2k+2]
=[(-1)^k](k+1)(k+2)=R.H.S.

2012-04-11 15:42:08 補充：
(k+1)(1)+(k)(2)+(k-1)(3)+...+3(k-1)+2(k)+1(k+1)
=k(1)+1+(k-1)(2)+2+(k-2)(3)+3+...+2(k-1)+(k-1)+(k)+k+(k+1)
=k(1)+(k-1)(2)+(k-2)(3)+...+3(k-2)+2(k-1)+1(k)+1+2+3+...+k+1
=k(k+1)(k+2)/6+1+2+3+...+k+1

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[匿名]

2012-04-11 11:33 pm

im not clever enough, could show more steps pls.

L.H.S.=(k+1)(1)+(k)(2)+(k-1)(3)+...+3(k-1)+2(k)+1(k+1)
? step
?step
=k(k+1)(k+2)/6+1+2+3+...+k+1

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