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Prove that 1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n terms = n/2 (9n^3 + 4n^2 - 4n - 1)

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Your problem `->` Prove that 1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n terms = n/2 (9n^3 + 4n^2 - 4n - 1)


L.H.S. `= 1 × 2^2 + 3 × 5^2 + 5 × 8^2 + ... n` terms

`= sum [ f(n) ]`

`= sum [ (2n - 1)(3n - 1)^2 ]`

`= sum [ (2n - 1)(9n^2 - 6n + 1)]`

`= sum [ 18n^3 - 21n^2 + 8n - 1]`

`= 18 sum n^3 - 21 sum n^2 + 8 sum n - sum 1`

`= 18 * (n^2 (n+1)^2)/4 - 21 * (n (n + 1) (2n + 1))/6 + 8 * (n (n+1))/2 - n`

`= n/2 [ 9 n (n + 1)^2 - 7(n + 1) (2n + 1) + 8 (n+1) - 2 ]`

`= n/2 [ 9 n (n^2 + 2n +1) - 7 (2n^2 + 3n + 1) + 8 n + 8 - 2 ]`

`= n/2 [ 9 n^3 + 18 n^2 + 9n - 14n^2 - 21n - 7 + 8 n + 8 - 2 ]`

`= n/2 [ 9 n^3 + 4 n^2 - 4n - 1 ]`

`=` R.H.S. (Proved)






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