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

Let `P9n):1.3+3.5+5.7+....+(2n-1)(2n+1)`
`(n(4n^2+6n-1))/(3)`
Step I For `n=1`.
LHS of Eq. (i) `=1.3=3`
RHS of Eq. (i) `=(1(4xx1^2+6xx1-1))/(3)=(4+6-1)/(3)=3`
`therefore` LHS = RHS
Therefore , P(1) is true.
Step II Assume that the result is true for `n=k`. Then, `P(k):1.3+3.5+5.7+.....+(2k-1)(2k+1)=(k(4k^2+6k-1))/(3)`
Step III For `n=k+1`, we have to prove that `P(k+1):1.3+3.5+5.7+....+(2k-1)(2k+1)+(2k+1)(2k+3)`
`=((k+1)[4(k+1)^2+6(k+1)-1])/(3)`
`=((k+1)(4k^2+14k+9))/(3)`
LHS `=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2k+1)(2k+3)`
`=(k(4k^2+6k-1))/(3)+(2k+1)(2k+3)` [by assumpation step ]
`=(4k^3+6k^2-k)/(3)+(4k^2+8k+3)`
`=(4k^3+18k^2+23k+9)/(3)`
`=((k+1)(4k^2+14k+9))/(3)=RHS`
This shows that the result is true for `n=k+1`. Therefore , by the principle of mathematical induction , the result is true for all `n in N`.