Hi Guys, i am almost there with this question but can't quite get the last step!!!
Use mathematical induction to prove that for all positive integers n:
1
2+3
2+5
2+...+(2n-1)
2= (1/3)n(2n-1)(2n+1)
Thanks!
Sk= (1/3)k(2k-1)(2k+1)
We must show that Sk+1 = (1/3)(k+1) ( 2(k+1) -1) (2(k+1) +1) = (1/3) (k+1)(2k+1)(2k+3)
Now, Sk+1 = Sk+Tk+1
Sk+1 = [ (1/3)k(2k-1)(2k+1) ] + [ (2(k+1) -1 )^2 ]
= (1/3)k(2k-1)(2k+1) + (2k+1)^2
= (2k+1) [ (1/3)k(2k-1) + (2k+1) ]
= (2k+1) [ (1/3) [2k^2 -k] +2k+1 ]
=(2k+1) [ (2/3)k^2 +(5/3)k +1 ]
= (1/3) (2k+1) [ 2k^2 +5k +3]
= (1/3) (2k+1) (2k^2 +2k +3k +3)
= (1/3) (2k+1) ( 2k(k+1) +3(k+1) )
= (1/3) (2k+1) (2k+3)(k+1)
As required.