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Geometric Progression 



Problem 13 of 23 


13. Prove that `1 * (2^2  3^2) + 2 * (3^2  4^2) + 3 * (4^2  5^2) + ... n` terms `= n/6 (n + 1) (4n + 11)`










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Geometric Progression 
13. Prove that `1 * (2^2  3^2) + 2 * (3^2  4^2) + 3 * (4^2  5^2) + ... n` terms `= n/6 (n + 1) (4n + 11)`
L.H.S. `= 1 × (2^2  3^2) + 2 × (3^2  4^2) + 3 × (4^2  5^2) + ... n` terms
`= sum [ f(n) ]`
`= sum [ n ((n + 1)^2  (n + 2)^2) ]`
`= sum [ n (n^2 + 2n + 1  n^2  2n  4) ]`
`= sum [ n (2n  3) ]`
`= sum (2n^2  3n) ]`
`= 2 sum n^2  3 sum n`
`= 2 * (n (n + 1) (2n + 1))/6  3 * (n (n + 1))/2`
`=  n/6 (n + 1) [ 2(2n + 1) + 9 ]`
`=  n/6 (n + 1) (4n + 11)`
`=` R.H.S. (Proved)





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