# How to rewrite a Sum?

How to find the following sum? $$\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right)$$ For example $$\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$$ I need to find the value of which this would converge to and I know that it must be six because the integral from $1$ to $3$ of the function $(2x-1)$ is $6$.

Any help is appreciated!!!

• $\sum i = (n)(n+1)/2$ ? – King Tut May 15 '18 at 5:38
• A couple of hints: A constant can always be moved in front of the summation, like this: $$\sum_n {a f(n)} = a \sum_n f(n)$$ And the sum is also distributive: $$\sum_{n} (a_n + b_n) = \sum_n a_n + \sum_n b_n$$ – Matti P. May 15 '18 at 5:38
• @Matti Infinite sum also? distributive? – King Tut May 15 '18 at 5:39
• Sorry, I meant in a finite case, like this. – Matti P. May 15 '18 at 5:40

Hint: You have to use $$\sum_{i=1}^n (a_i+b_i) = \left(\sum_{i=1}^na_i\right)+\left(\sum_{i=1}^n b_i\right)$$ and $$\sum_{i=1}^n (\lambda a_i)=\lambda\left(\sum_{i=1}^n a_i\right)$$ For the second equation, you have to check that $\lambda$ can not depend on $i$ but it can depend on other variables like $n$.
$$\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) = \frac{2}{n}\left( \sum_{i=1}^{n} \frac{4i}{n} + \sum_{i=1}^{n} 1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i + \frac{2}{n}\cdot n = \frac{8}{n^2}\cdot \frac{n(n+1)}{2} + 2 = 4 \left(1+\frac{1}{n} \right) + 2$$