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Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)

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Your problem `->` Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)


L.H.S. `= 2 + 5 + 10 + 17 + ... n` terms

`= sum [ 1 + n^2 ]`

`= sum 1 + sum n^2`

`= n + (n (n + 1) (2n + 1))/6`

`= n/6 [ 6 + (2n^2 + 3n + 1) ]`

`= n/6 (2n^2 + 3n + 7)`

`=` R.H.S. (Proved)






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