A.Maths (MI) 趕趕趕!!!!

a. 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6
1+2+3+4....+n=n(n +1)/2
Let P(n) = 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6
When n = 1
L.H.S = 1x5 = 5
R.H.S = n(n+1)(2n+13)/6 = 1(1+1)(2x1+13)/6 = 5
As L.H.S = R.H.S
So P(1) is true
Assume P(k) is true, i.e. 1x5+2x6+3x7+....+k(k+4)= k(k+1)(2k+13)/6
When n = k+1
L.H.S.
= 1x5+2x6+3x7+....+k(k+4)+(k+1)(k+5)
= k(k+1)(2k+13)/6+(k+1)(k+5)
= k(k+1)(2k+13)/6+6(k+1)(k+5)/6
= (k+1)[k(2k+13)+6(k+5)]/6
= (k+1)[2k2+13k+6k+30]/6
= (k+1)[2k2+19k+30]/6
= (k+1)(k+2)(2k+15)/6
=R.H.S.
So P(k+1) is true.
By Mathematical Induction, 1x5+2x6+3x7+....+n(n +4)= n(n+1)(2n+13)/6 is true for all positive integer of n

find 1x4+2x5+3x6+.....+n( n+3)
1x(5-1)+2x(6-1)+3x(7-1)+.....+n( n+4-1)
=1x5 + 2x6 + 3x7 + ……+n(n+4) -1-2-3-….-n
= n(n+1)(2n+13)/6 - n(n +1)/2
= n(n+1)(2n+13)/6 - 3n(n +1)/6
= n(n+1)[(2n+13) – 3]/6
= n(n+1)(2n+10)/6
= n(n+1)(n+5)/3

and 12+22+......+n2
= 1x1+1x4+2x2+2x4+3x3+3x4+…..+nxn+nx4–1x4–2x4–3x4-….-nx4
= 1x5+2x6+3x7+…..+n(n+4)–4(1+2+3+….+n)
= n(n+1)(2n+13)/6–4n(n +1)/2
= n(n+1)(2n+13)/6–12n(n +1)/6
= n(n+1)[(2n+13)–12]/6
= n(n+1)[2n+13–12]/6
= n(n+1)(2n+1)/6