`1.3+3.5+5.7+......+(2n-1)(2n+1)=(n(4n^2+6n-1))/3`

Explanation

Let
`P(n) :1 .3+3.5+5.7 +……+ (2n -1)(2n+1)`
`=1/3 n(4n^(2)+6n-1)`
For n =1
`L.H.S. =1.3=3`
`R.H.S.=1/3 .1.(4.1^(2)+6.1-1)`
`=1/3(4 +6-1)=3`
`:. L.H.S. =R.H.S.`
Therefore P (n) is true for n=1
Let P (n) be true for n=k.
`p(k) : 1.3 +3.5 +5.7 +....+(2k-1)(2k+1)`
`=1/3 k(4k^(2)+6k-1)`
For n=k+1
` 1.3 +3.5+5.7 +.......`
`+(2K-1)(2K+1)+(2k+1)(2K+3)`
`=1/3k(4K^(2)+6K-1)+(2K+1)(2K+1)(2K+3)`
`(K(4K^(2)+6K-1)+3(2K+1)(2K+3))/(3)`
`=1/3(4k^(2)+18K^(2)+23k+9)`
`=1/3(k+1)(4k^(2)+14k+9)`
`=1/3(k+1)[4(K^(2)+2K+1)+6(K+1)-1]`
`=1/3(k+1)[4(K+1)^(2)+6(K+1)-1]`
`rArr` P (n) is also true for n=K+1
Hence from the principle of mathematical induction P (n) is true for all positive integers n .