Solving a summation using summation rules

Question

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

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

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 '19 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 '19 at 18:59
... or $n+1$ of them, depending on the lower limit (which I missed - sorry). — NickD, Jan 24 '19 at 18:59

score 2 · Accepted Answer · edited Feb 22 '22 at 13:19

If the OP has parenthesized things correctly in the question, it should go like this: \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*}

Solving a summation using summation rules

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