Question

How do you solve `\sum_{k=1}^2 (3-k)`?

Answer 1

`sum_(k=1)^2(3-k) = 3`

Explanation:

This problem is based around understanding sigma notation. In general, given a function `f` and integers `a, b` where `a<=b`, we have

`sum_(k=a)^bf(k) = f(a)+f(a+1)+f(a+2)+...+f(b)`

that is, we look at the sum of `f`, evaluated at each integer from (and including) `a` to `b`.

In this case, our function is `f(x) = 3-x`, our `a` is `1`, and our `b` is `2`. Expanding, we get our result:

`sum_(k=1)^2(3-k)= (3-1)+(3-2)`

`= 2+1`

`= 3`