Help with Algebra II question.

Question

cmadaviss · Answer

$$\sum_{k=0}^{1}(2-k)= ?$$

I need help solving this equation!

hartnn · Answer

you know how $\sum $ operator works ?

cmadaviss · Answer

No we haven't learned this yet, but it's on a bonus assignment

hartnn · Answer

ok, 
so we insert all the values of k from its lower limit to the upper limit in the given function,

for example:

hartnn · Answer

$\Large \sum \limits_{k=a}^{a+n}f(k) = f(a) + f(a+1) + f(a+2) + ... f(a+n) $

hartnn · Answer

so in your problem,

you'll first put k= 0 in 2-k

then you'll put k = 1 in 2-k

pause here,
and then add all the above answers

hartnn · Answer

do you understand this?

cmadaviss · Answer

I plug both o and 1 into my equation?

hartnn · Answer

yes, 
like this :
$\Large \sum \limits_{k=0}^1 (2-k)= (2-0) + (2-1)  = ...$

cmadaviss · Answer

And I solve from there?

hartnn · Answer

yup

cmadaviss · Answer

so k=-1 ?

hartnn · Answer

how?

also we are not trying to find the value of k

hartnn · Answer

we are plugging in different values of k

here, from 0 to 1
so k =0 and then k =1

hartnn · Answer

in the given function 2-k

and then add the results, because thats what the $\sum $ operator mean....summing of all the fuanction values

cmadaviss · Answer

Oh okay

cmadaviss · Answer

SO it's 3?

hartnn · Answer

yes, correct!
$\huge \checkmark $

cmadaviss · Answer

Thank you!

hartnn · Answer

welcome ^_^