Prove by using the principle of mathemtical induction: ` 1^3+3^3+5^2+…+(2n-1)^2 = (n(2n-1)(2n+1))/3 `

Explanation

`"Let "P (n) : 1^(2)+3^(2)+5^(2)+......+(2n-1)^(2)`
` =1/3 n(4n^(2)-1)`
for n =1
`L.H.S. =1^(2) =1, R.H.S. =1/3 .1.(4.1^(2)-1)=1`
`:. L.H.S. =R.H.S.`
`rArr` P(n) is true for n=1
Let P(n) be true for n=K
`:. P(k) : 1^(2) +3^(2)+5^(2)+`
`.......+(2k-1)^(2)=1/3 k(4k^(2)-1)`
for n=K+1
`P(k+1) : 1^(2) +3^(2) +5^(2)+......+(2K-1)^(2)`
`+(2K+1)^(2)=1/3 k(4k^(2)-1)+(2k+1)^(2)`
`(k(4k^(2)-1)+3(2k+1)^(2))/(3)`
`=1/3[4k^(3)-k+12k^(2)+12k+3]`
`=1/3[4k^(3)+12k^(2)+11k+3]`
`=1/3(k+1)(4k^(2)+8k+3)`
`=1/3(k_+1)[4(k^(2)+2k+1)-1]`
`=1/3(k+1)[4(k+1)^(2)-1]`
`rArr` P (n) is also true for n=K+1
Hence from the principle of mathematical induction P (n) is true for all natural numbers. n