0
$\begingroup$

So I need help computing the following sum.

Question: Suppose $ x_1 = 5, x_2 = −3, x_3 = 7, x_4 = 8,$ and $x_5 = 2$. Compute the following sum.

$(\sum_{i=1}^5 x_i)+10$ where the upper limit is 5, and the lower limit is $i = 1$. The expression is $(x_i) + 10$. keep in mind the $+10$ is outside the brackets.

What I think:

Since we are given the values for each $x$ $(5,-3,7,8,2)$, i am assuming that it would be $(5+10) + (-3+10)$ so and so forth.

Now that is what I think. I don't know the correct answer. So can someone please help me figure it out.

Thank You.

$\endgroup$
  • 1
    $\begingroup$ Addition does not distribute like that ($x+(y+z) \neq (x+y) + (x+z)$). The parentheses are emphasizing that $\sum_{i=1}^5 x_i$ should be computed first and then to that we add $10$. What is the value of $\sum_{i=1}^5 x_i$ going to be? $\endgroup$ – Trevor Norton Feb 26 '17 at 18:26
  • $\begingroup$ In the edit you added another question,you shouldn't do that,anyway I would suggest you to try it yourself and tell us where you are stuck(preferably in another question). $\endgroup$ – kingW3 Feb 26 '17 at 19:16
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. Ref: meta.math.stackexchange.com/a/20529/290189 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 27 '17 at 22:06
4
$\begingroup$

As you say, the addition of $10$ is outside the brackets. So you make the sum first:

$(\sum_{i=1}^5 x_i)+10= (5+-3+7+8+2)+10 = 19+10=29$

$\endgroup$
4
$\begingroup$

Be careful, as you said, the $+10$ is outside of the brackets: $$(\sum_{i=1}^{5}x_{i})+10.$$

Since the order of operations tell us to compute expressions within parenthesis first, we want to first compute the sum $\displaystyle\sum_{i=1}^{5}x_{i}$ and then add $10$ to the result.

On the other hand, if we were given $$\sum_{i=1}^{5}(x_{i}+10)$$ then we would have the following sum $$(x_{1}+10)+(x_{2}+10)+(x_{3}+10)+(x_{4}+10)+(x_{5}+10).$$ Here, the parenthesis are implying that the $+10$ is in each term of the sum.

$\endgroup$
  • 1
    $\begingroup$ Thank You very much but a quick question, If there was a Square outside the brackets, then we would still do the sum first then the Square of the total sum. Correct?/? $\endgroup$ – Jainam Patel Feb 26 '17 at 18:32
  • $\begingroup$ Yes, if you had $$\big[\sum_{i=1}^{5}x_{i}\big]^{2},$$ you would indeed compute $\sum_{i=1}^{5}x_{i}$ first, and then square the result. $\endgroup$ – yung_Pabs Feb 26 '17 at 18:44
  • $\begingroup$ Hey Can you help me with this. can you check the updated question. please and thank you. $\endgroup$ – Jainam Patel Feb 26 '17 at 19:00
  • $\begingroup$ @JainamPatel No!!! See this post about question edit on meta.mathse by quid: "As a rule a question should not be a moving target, and to alter it in such a way as to render existing answers wrong or even just incomplete is discouraged. It is alright to rollback such edits and to ask OP to ask a new question." $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 27 '17 at 22:28

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.