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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)

Solution:
Your problem `->` 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)






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