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

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	Problem 19 of 23

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

SolutionExample

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Geometric Progression

19. 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)