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Geometric Progression 



Problem 18 of 23 


18. Prove that `1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n` terms `= n/12 (n + 1)^2 (n + 2)`










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Geometric Progression 
18. Prove that `1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n` terms `= n/12 (n + 1)^2 (n + 2)`
L.H.S. `= 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n` terms
`= sum [ f(n) ]`
`= sum [ sum n^2 ]`
`= sum [ n/6 (n + 1) (2n + 1) ]`
`= sum [ n/6 (2n^2 + 3n + 1) ]`
`= sum [ 1/6 (2n^3 + 3n^2 + n) ]`
`= 1/6 [ 2 sum n^3 + 3 sum n^2 + sum n ]`
`= 1/6 [ 2 * (n^2 (n+1)^2)/4 + 3 * (n (n + 1) (2n + 1))/6 + (n (n + 1))/2 ]`
`= 1/6 [ (n^2 (n+1)^2)/2 + (n (n + 1) (2n + 1))/2 + (n (n + 1))/2 ]`
`= 1/6 * (n (n + 1))/2 [ n (n+1) + (2n + 1) + 1 ]`
`= n/12 (n + 1) [ n^2 + n + 2n + 2 ]`
`= n/12 (n + 1) [ n^2 + 3n + 2 ]`
`= n/12 (n + 1) [ (n + 2) (n + 1) ]`
`= n/12 (n + 1)^2 (n + 2)`
`=` R.H.S. (Proved)





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