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Use induction to prove that
\sum _{i=1}^n \left(3 i^2+i\right) = n (n+1)^2.
An archived instance of a Discrete forum
Another summation induction
ksimmon1
Proof by induction:
\sum _{i=1}^n \left(3 i^2+i\right) = n (n+1)^2.
Base Case:
(3 i^2+i) = n (n+1)^2
(3 (1)^2+(1)) = (1) (1+1)^2.
4 = 4
Assume that for a fixed value of n that:
\sum _{i=1}^n \left(3 i^2+i\right) = n (n+1)^2
We want to show that:
\sum _{i=1}^{n+1} \left(3 i^2+i\right) = (n+1)(n+2)^2
Well,
\sum _{i=1}^{n+1} = \sum _{i=1}^{n} + 3(n+1)^2+1
= n(n+1)^2 + 3(n+1)^2+1
=(n+1)(n(n+1)+3(n+1)+1)
=(n+1)(n^2+4n+4)
=(n+1)(n+2)^2