Mathematic induction

1(a) Prove, P(n)
1*3+3*5+5*7+...+(2n－1)(2n+1)=[(2n－1)(2n+1)(2n+3)+3]/6
Sol
when n=1
L.H.S.=(2*1－1)*(2*1+1)=1*3=3
R.H.S.=[(2*1－1)*(2*1+1)(2*1+3)+3]/6=[1*3*5+3]/6=3
L.H.S.=R.H.S.
So, P(1) irs true

Assume that P(k) is true

1*3+3*5+5*7+...+(2k－1)(2k+1)=[(2k－1)(2k+1)(2k+3)+3]/6
So
1*3+3*5+5*7+...+(2k－1)(2k+1)+(2k+1)(2k+3)
=[(2k－1)(2k+1)(2k+3)+3]/6+(2k+1)(2k+3)
=(2k－1)(2k+1)(2k+3)/6+(2k+1)(2k+3)+3/6
=[(2k+1)(2k+3)/6]*[(2k－1)+6]+3/6
=[(2k+1)(2k+3)/6]*(2k+5)+3/6
=(2k+1)(2k+3)(2k+5)/6+3/6
=[2k(k+1)－1]*[2(k+1)+1]*[2(k+1)+3]/6+3/6
=｛[2k(k+1)－1]*[2(k+1)+1]*[2(k+1)+3]+3/6｝
So P(k+1) is true.(b) Using the result of (a),prove that,
1^2+2^2+3^2+...+n^2=[n(n+1)(2n+1)]/6Sol1*3+3*5+5*7+...+(2n－1)(2n+1)=[(2n－1)(2n+1)(2n+3)+3]/6

1*3+3*5+5*7+...+(2n－1)(2n+1)
=(2-1)*(2+1)+(4-1)*(4+1)+…+(2n-1)*(2n+1)
=(2^2-1)+(4^2-1)+…+(4n^2-1)
=(2^2+4^2+6^2+…+4n^2)-n
=4*(1^2+2^2+3^2+…+n^2)-n

4*(1^2+2^2+3^2+…+n^2)-n=[(2n－1)(2n+1)(2n+3)+3]/6
4*(1^2+2^2+3^2+…+n^2)=n+[(2n－1)(2n+1)(2n+3)+3]/6

24*(1^2+2^2+3^2+…+n^2)
=6n+(2n－1)(2n+1)(2n+3)+3
=6n+(8n^3+12n^2－2n－3)+3
=8n^3+12n^2+4n
=4n(2n^2+3n+1)
=4n(n+1)(2n+1)

(1^2+2^2+3^2+…+n^2) =n(n+1)(2n+1)/6

2(a)(i) Prove P(n): 1/2+1+3/2+...+n/2=n(n+1)/4
when n=1,
L.H.S.=1/2
R.H.S.=2/4=1/2
P(1) is true.

Assume that P(k) is true. i.e. 1/2+1+3/2+...+k/2=k(k+1)/4

when n=k+1
L.H.S.
=1/2+1+3/2+...+k/2+(k+1)/2
=k(k+1)/4+(k+1)/2
=[k(k+1)+2(k+1)]/4
=(k+1)(k+2)/4
=R.H.S.

So, P(k+1) is true. By M.I. P(n) is true for all posivitive integers n

(ii) Prove P(n)
1*5+2*7+...+n(2n+3)=[n(n+1)(4n+11)]/6

when n=1:
L.H.S.=5
R.H.S.=2*15/6=5
P(1) is true.

Assume that P(k) is true. i.e. 1*5+2*7+...+k(2k+3)=[k(k+1)(4k+11)]/6

when n=k+1
L.H.S.
=1*5+2*7+...+k(2k+3)+(k+1)(2k+5)
=[k(k+1)(4k+11)]/6+(k+1)(2k+5)
=[k(k+1)(4k+11)+6(k+1)(2k+5)]/6
=[(k+1)[(4k^2+11k)+(12k+30)]/6
=[(k+1)(4k^2+23k+30)]/6
=[(k+1)(k+2)(4k+15)]/6
=R.H.S.
By M.I. P(n) is true for all posivitive integers n

(b) 1*4+2*6+...+n(2n+2)
=1*(5-1)+2*(7-1)+....+n(2n+3-1)
=(1*5+2*7+...+n(2n+3))-(1+2+...+n)
=[n(n+1)(4n+11)]/6-n(n+1)/2
=[n(n+1)(4n+11)-3n(n+1)]/6
=[n(n+1)(4n+8)]/6
=2n(n+1)(n+2)/3