A.maths (urgent)

Solution:

圖片參考：http://hk.geocities.com/stevieg_1023/po1.gif


圖片參考：http://hk.geocities.com/stevieg_1023/po2.gif

Let S(n) be the statement " 1²+2²+…+n²=[n(n+1)(2n+1)]/6 "

When n=1,
L.H.S.= 1² =1
R.H.S.= [1(1+1)(2+1)]/6 = 1
so S(1) is true.

Assume S(k) is true,
i.e. 1²+2²+…+k²=[k(k+1)(2k+1)]/6

When n=k+1
R.H.S.=(k+1)(k+2)(2k+3)/6
L.H.S.=1²+2²+…+k²+(k+1)²
=[k(k+1)(2k+1)]/6+(k+1)²
=(k+1) [k(2k+1)+6k+6]/6
=(k+1)(2k²+7k+6)/6
=(k+1)(k+2)(2k+3)/6
=R.H.S.
so S(k+1) is true.
By M.I. , S(n) is true for any positive integer n.


3²+6²+…+99²
=1²(3²)+2²(3²)+…+33²(3²)
=3²(1²+2²+…+33²)
=9([33(33+1)(66+1)]/6
=3(33x17x67)
=112761

[3(3+1)(2x3+1]/6 -[2(2+1)(2x2+1)/6
=(3x4x7)/6-(2x3x5)/6
=84/6-30/6
=14-5
=9