中4,a-maths問題...

As follow AS~~~

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對於所有正正數n :1．４＋２．７＋３．１０＋．．．．＋ｎ（３ｎ＋１）＝ｎ（ｎ＋
１）^2


設P(n)為這個命題"1．４＋２．７＋３．１０＋．．．．＋ｎ（３ｎ＋１）＝ｎ（ｎ＋
１）^2"

當n=1

左方=(3+1) 右方=(1+1)^2
=4 =4

∵左方=右方
∴p(1)成立

設p(k)成立,K是正整數

即1．４＋２．７＋３．１０＋．．．．＋k（３k＋１）＝k（k＋
１）^2


當n=k+1

左方=．４＋２．７＋３．１０＋．．．．＋k（３k＋１）+(k+1)(3k+4)

=k（k＋１）^2+(k+1)(3k+4)
=(k+1)[k(k+1)+(3k+4)]
=(k+1)(k^2+k+3k+4)
=(k+1)(K^2+4K+4)
=(k+1)(k+2)^2

右方=(k+1)(k+2)^2



∵p(k)成立,則p(k+1)都成立
根據數學歸納法,
對於所有正整數n"1．４＋２．７＋３．１０＋．．．．＋ｎ（３ｎ＋１）＝ｎ（ｎ＋
１）^2"

P(n)成立

Assume 1*4+2*7+3*10+...+k(3k+1)=k(k+1)^2 , where k is a positive integer

Consider when n=k+1
1*4+2*7+...+k(3k+1)+(k+1)(3k+4)
=k(k+1)^2+(3k+4)(k+1)
=(k+1)[k(k+1)+(3k+4)]
=(k+1)(k^2+4k+4)
=(k+1)(k+2)^2//

So it is true for n=k+1
By Mathematical Induction , it is true for all positive integer n.