F. 6 Mathematical Induction

As follows:


圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/01-165.jpg

To prove that for any positive integer n,

1x2 + 2x3 + 3x4 + ... + (3n)(3n+1) = n(3n+1)(3n+2)

Method:

When n=1, LHS equals
1x2 + 2x3 + 3x4 = 20

RHS equals
1(3+1)(3+2) = 20

Therefore it is true for n=1

Assume that it is true for n=k.
That is: 1x2 + 2x3 + 3x4 + ... + (3k)(3k+1) = k(3k+1)(3k+2)

When n = (k+1)

LHS equals
[ 1x2 + 2x3 + 3x4 + ........... + (3k)(3k+1) ] + (3k+1)(3k+2) + (3k+2)(3k+3) +
[3(k+1)][3(k+1) + 1]
=[k(3k+1)(3k+2)] + [3k+1][3k+2] + [3k+2][3k+3] + [3k+3][3k+4]
=[k(9k^2 + 9k + 2)] + [9k^2 + 9k + 2] + [9k^2 + 15k + 6] + [9k^2 + 21k + 12]
=[9k^3 + 9k^2 + 2k] + [27k^2 + 45k + 20]
=[9k^3 + 36k^2 + 47k + 20]
=(k+1)(9k^2 + 27k + 20)
=(k+1)(3k+4)(3k+5)

RHS equals
(k+1)[ 3(k+1) + 1 ][ 3(k+1) + 2 ] = (k+1)(3k+4)(3k+5)

Therefore it is also true when n = k+1

Since it is true for n=1; and, when it is true for n=k it is also true for n=(k+1);
by mathematical induction, it is true for all positive integer n.

The question do not confine the method, so

1x2 + 2x3 + 3x4 + ........... + (3n)(3n+1)

　3n
= Σ k(k + 1)
　k=1

= Σ k^2 + Σ k
...

Or
Σ k(k + 1)

　3n
= 1/3[ Σ k(k + 1)(k + 2) - Σ (k - 1)k(k + 1) ]
　k=1
...