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

We have,
`P(n) : 1 xx 3 + 3 xx 5 + 5 xx 7 + "….." + (2n-1) xx (2n+1)`
`= (n(4n^(2) + 6n - 1))/(3)`
For `n = 1`,
L.H.S. `= 1 xx 3 = 3`,
R.H.S. `= (1.(4+6-1))/(3) = 3`
Thus, `P(1)` is holds.
Let `P(n)` be true for some `n = k`.
`1 xx 3 + 3 xx 5 + 5 xx 7 + "......"+ (2k-1)(2k+1)+(2k+1)(2k+3)`
`= (k(4k^(2) + 6k - 1))/(3) + (2k + 1)(2k+3)`
`= (k(4k^(2) + 6k - 1) + 3(4k^(2) + 8k + 3))/(3)`
`= (4k^(3) + 18k^(2) + 23k + 9 )/(3)`
`= ((k+1)(4k^(2) + 14 k + 9))/(3)`
`= ((k+1){4(k+1)^(2) + 6(k+1)-1})/(3)`
Thus, `P(k+1)` is true whenever `P(k)` is true.
Hence by the principle of mathematical induction, statement `P(n)` is true for all natural number.