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 $n$.

confused on one question The question implies that you must have seen something at least remotely similar in the past. What's different here, what did you try, and where did you get stuck? — dxiv, Sep 19 '18 at 06:11
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Martin Sleziak, Sep 19 '18 at 06:13
@jessica.m Since you are new here, recall also that you can accept an answer among the given, more details here LINK — user, Sep 19 '18 at 17:50

user · Answer 1 · 2018-09-19T06:16:26.443

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?

edited Sep 19 '18 at 06:16

answered Sep 19 '18 at 06:09

user

154,566

Really thank you sooo much, very helpful. – jessica.m Sep 19 '18 at 17:27
You are welcome! Bye – user Sep 19 '18 at 17:29

score 1 · Answer 2 · answered Sep 19 '18 at 06:35

1

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

solve it to get your answer.

answered Sep 19 '18 at 06:35

NewBornMATH

772

Really thank you sooo much, very helpful. – jessica.m Sep 19 '18 at 17:27

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

2 Answers2