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Geometric Progression 



Problem 16 of 23 


16. Prove that for all n belongs to N, 1^{2} × n + 2^{2} (n  1) + 3^{2} (n  2) + ... + n^{2} × 1 = ^{n}/_{12} (n + 1)^{2} (n + 2)










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Geometric Progression 
16. Prove that for all n belongs to N, 1^{2} × n + 2^{2} (n  1) + 3^{2} (n  2) + ... + n^{2} × 1 = ^{n}/_{12} (n + 1)^{2} (n + 2)
Here, `f(r) = r^2 (n r + 1) = (n + 1) r^2  r^3`
Now, L.H.S. `= 1^2 × n + 2^2 (n  1) + 3^2 (n  2) + ... + n^2 × 1`
`= sum [ (n + 1) r^2  r^3 ]` (Where r = 1 to n)
`= (n + 1) sum r^2  sum r^3` (Where r = 1 to n)
`= (n + 1) * (n (n + 1) (2n + 1))/6  (n^2 (n + 1)^2)/4`
`= (n (n + 1)^2 [ 2 (2n + 1)  3 n])/12`
`= (n (n + 1)^2 (n + 2))/12`
`=` R.H.S. (Proved)







