0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ @nickD the starting point of the lower limit is different $\endgroup$ – Arthur Green Jan 24 at 18:57
  • 1
    $\begingroup$ Yes to the last prompt. The second and third terms are independent of $i$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 24 at 18:57
  • 1
    $\begingroup$ 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. $\endgroup$ – NickD Jan 24 at 18:59
  • $\begingroup$ ... or $n+1$ of them, depending on the lower limit (which I missed - sorry). $\endgroup$ – NickD Jan 24 at 18:59
  • 2
    $\begingroup$ Hint: $\sum_{i=1}^n 1 = n$. $\endgroup$ – NickD Jan 24 at 19:01
2
$\begingroup$

If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. 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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.