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