## anonymous 3 years ago Finding the indicated sum ,

1. anonymous

2. whpalmer4

This one can be solved without any cleverness. $\sum_{k=1}^{4}(-1)^k(k+11) =$$(-1)^1(1+11)+(-1)^2)(2+11)+(-1)^3)(3+11)+(-1)^4(4+11) =$

3. whpalmer4

Now why did I stick in those unwanted )'s? $(-1)^1(1+11)+(-1)^2(2+11)+(-1)^3(3+11)+(-1)^4(4+11) =$ For odd k, $$(-1)^k = -1$$ For even k, $$(-1)^k = 1$$ -(1+11)+(2+11)-(3+11)+(4+11) = all the alternating of signs cancels out the 11s, so it's just -1 + 2 - 3 + 4 = 2

4. anonymous

How did you get the -1+2-3+4?

5. anonymous

When I solved it I didnt get none of that, maybe I did it wrong?

6. whpalmer4

$-(1+11)+(2+11)-(3+11)+(4+11) = -1 -11 + 2 + 11 - 3 - 11 + 4 + 11$ The alternating -11 + 11 sum to 0, leaving -1 + 2 - 3 + 4

7. anonymous

Ah okay, I get it