Summation of simple series $\sum_{r=5}^n (2r + 4r^2) $

Question

I am trying to teach myself, but I am confused on one question. It says "for the following summation, give an equivalent equation without the summation:

$$\sum_{r=5}^n (2r + 4r^2) $$ where $i$ takes values from $5$ to $n$.

Answer 1

HINT

Note that

$$\sum 2i+4i^2=2\sum i+4\sum i^2$$

and

$$\sum_5^n a_n=\sum_1^n a_n-\sum_1^4 a_n$$

then refer to the sum of powers formulas.

Refer also to the related How Are the Solutions for Finite Sums of Natural Numbers Derived?

Answer 2

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Seems to be a high school level problem : $$\sum_{r=5}^{n}2r+r^2=2\sum_{r=5}^{n}r+\sum_{r=5}^{n}r^2 $$ $$=2\bigg(\sum_{r=1}^{n}r-\sum_{r=1}^{4}r\bigg)+\bigg(\sum_{r=1}^{n}r^2-\sum_{r=1}^{4}r^2\bigg)$$ $$=2\bigg(n-\frac{5\times4}{2}\bigg)+\bigg(\frac{(n^2+n)(2n+1)}{6}-9\frac{(4^2+4)}{6}\bigg)$$

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solve it to get your answer.