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Geometric Progression 



Problem 17 of 23 


17. 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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Geometric Progression 
17. 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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