Prove the following by the principle of mathematical induction: `\ 1. 3+2. 4+3. 5++(2n-1)(2n+1)=(n(4n^2+6n-1))/3`

Explanation

`P(n): 1.3+3.5+5.7+....+(2n-1)(2n+1)=(n(4n^(2)+6n-1))/(3),`
When n=1 L.H.S. 1.3=3
and R.H.S =`(1(4xx1^(2)+6xx1-1))/(3)`
and R.H.S. `=(1(4xx1^(2)+6xx1-1))/(3)`
`=(9)/(3)=3`
`therefore L.H.S=R.H.S ` Hence P(1) is true ....(a)
Let P(k) be true.
`therefore 1.3+3.5+5.7+......(2k+1)(2k+1)=(k.(4k^(2)+6k+1))/(3).....(i)`
To prove P(k+1) is true i.e.
`1.3+3.5+5.7+......(2k-1)(2k+1)+{2(k+1)-1} {2(k+1)+1}`
`=((k+1)[4(k+1)^(2)+6(k+1)-1])/(3).....(ii)`
"Adding" {2(k+1)-1} {2(k+1)+1} i.e. (2k+1)(2k+3) to both sides of (i), we get
[1.3+3.5+5.7.......+(2k+1)(2k+1)]+(2k+1)(2k+3)
`=(k(4k^(2)+6k-1))/(3)+(2k+1)(2k+3)`
`=(1)/(3)[4k^(3)+6k^(2)-k+3(4k^(2)+8k+3)]`
`=(1)/(3)[4k^(3)+6k^(2)-k+12k^(2)+24k+9]`
`=(1)/(3)[4k^(3)+18k^(2)+23k+9]`
and `((k+1)[4(k+1)^(2)+6(k+1)-1])/(3)`
`=((k+1)[4k^(2)+8k+4+6k+6-1])/(3)`
`=(1)/(3)[4k^(3)+18k^(2)+23k+9]`
Hence, P(k+1) is true whenever P(k) is true ....(b)
From (a) and (b) by the principle of mathematical induction it follows is true for all natural number.