# Solving a summation using summation rules

I am given two problems.

$$\sum_{i=0}^{n} (4i + \frac{3}{4}n + \frac{1}{2}) \tag 1$$

$$\sum_{i=1}^{n} (4i + \frac{3}{4}n + \frac{1}{2}) \tag 2$$

I am asked to solve it.

I know I can manipulate this into three separate summations:

Attempt:

$$\sum_{i=0}^{n} (4i) + \sum_{i=0}^{n} (\frac{3}{4}n) + \sum_{i=0}^{n} (\frac{1}{2}) \tag 1$$

Unfortunately, I am confused on how to proceed other than the fact that the third summation will turn into $$\ \frac{1}{2}$$. (because there is no i or n)

May anyone share with me the next step? I'm not sure in what form the solution should be in.

EDIT:

would the first term turn into: $$\ 4 \times \frac{n(n+1)}{2}$$

EDIT 2:

I've currently got

$$\left(4 \times \frac{n(n+1)}{2}\right) + (?) + \frac{1}{2}n \tag 2$$

• @nickD the starting point of the lower limit is different – Arthur Green Jan 24 at 18:57
• Yes to the last prompt. The second and third terms are independent of $i$. – GNUSupporter 8964民主女神 地下教會 Jan 24 at 18:57
• Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $\frac{1}{2}$: it is a sum of $n$ of them. – NickD Jan 24 at 18:59
• ... or $n+1$ of them, depending on the lower limit (which I missed - sorry). – NickD Jan 24 at 18:59
• Hint: $\sum_{i=1}^n 1 = n$. – NickD Jan 24 at 19:01

\begin{align*} \sum_{i=0}^{n} \left(4i + \frac{3}{4}n + \frac{1}{2}\right) & = \sum_{i=0}^{n} \left(4i \right) + \sum_{i=0}^{n}\frac{3}{4}n + \sum_{i=0}^{n}\frac{1}{2} \\ &= 4 \sum_{i=0}^{n} i + \frac{3}{4}n\sum_{i=0}^{n} 1 + \frac{1}{2}\sum_{i=0}^{n} 1\\ & = 4 \frac{n(n+1)}{2} + \frac{3}{4}n(n+1)+ \frac{1}{2}(n+1) \\ & = 2n (n + 1) + \frac{3}{4}n(n+1)+ \frac{1}{2}(n+1)\\ & = \frac{11}{4} n(n+1) + \frac{1}{2} (n+1) \end{align*}