How to compute the summation of f(x) when the limits are in terms of y?

Question

This is the relevant part of the question:

$\sum_{y=0}^3 (x^2)$

I know the answer is $4x^2$ but I'm not certain why that is.

I'm assuming that it is because there are four elements (i.e. 0,1,2 & 3) but I don't see why that would result in a value of $4x^2$.

If you assumed the summation to be over $0y.x^2$ then the answer would be zero.

Any clarification in this would be appreciated!

Answer 1

In general you have:

$$\sum_{y=a}^bf(y)=f(a)+f(a+1)+f(a+2)+\cdots+f(b-2)+f(b-1)+f(b)$$

In your sum there are four terms for $y\in\{0,1,2,3\}$ and $f(y)=x^2$ is a constant so you get:

$$\sum_{y=0}^{3}x^2=x^2+x^2+x^2+x^2=4x^2$$

Answer 2

When you substitute $y$ into $x^2,$ you get $x^2$ no matter what the value of $y,$ since $y$ doesn't appear in the formula. Therefore, you have a sum of $4$ terms, each of which is $x^2.$

Answer 3

Consider a (fairly useless) function of $y$ that always returns the same specific value $V$ no matter what value of $y$ you feed into it. For example, $$ w(5) = 5$$

If you were to sum that function over some range of $y$, then that value would sum to just $nV$ where $n$ is the number of terms in the sum. So for example, $$ \sum_{n=1}^6 5 = 6\cdot 5 = 30 $$ Your problem is the same zs this, only replacing the $V$ or the $5$ with $x^2$, which does not depend on $y$ at all.