Summation Notations - (Discrete Math) I'm having trouble

Question

I've been working on a single sigma notation problem for about 40 minutes now, and I'm having trouble conceptualizing the solution to this problem. I look at examples, and it seems that I begin to understand the process, and then I'm thrown for a loop once I try to integrate what I thought I learned.

$$\sum_{i=-1}^4(-i)^{i+1}$$ equation

Sorry for including an image, I couldn't figure out how to implement LaTeX.

For the increments, I had written down (-1 = 0) (0 = 0) (1 = -1) (2 = 8) (3 = -81) (4 = -1024)

Any help is appreciated, I am really struggling and this is one of the few times in my life I've attempted to get help.

Thank you.

Answer 1

I'm afraid I'm not sure what you mean when you write "for the increments..."

When $i=-1$, the summand is $(-(-1))^{-1+1}=1^0=1$.

When $i=0$, the summand is $(-0)^{0+1}=0^1=0$.

When $i=1$, the summand is $(-1)^{1+1}=(-1)^2=1$.

When $i=2$, the summand is $(-2)^{2+1}=(-2)^3=-8$.

And so on. You'll find that the sum evaluates to $-949$.

Answer 2

Your particular sum is a bit knotty, but here are some common patterns that may help you in evaluating such things, or just having some intuition about how they work:

$$\sum_{i=1}^{5} (-1)^i = -1 + 1 -1 + -1 + 1$$

Notice that the even terms are always positive, and the odd terms are negative. This is a rather common pattern in sums.

We also have

$$\sum_{i=-3}^{3} i^n = (-3)^n + (-2)^n + (-1)^n + 0 + 1^n + 2^n + 3^n \space \space \text{(for odd $n$)}$$

$$\sum_{i=-3}^{3} i^n = 3^n + 2^n + 1^n + 0 + 1^n + 2^n + 3^n = 2(1^n + 2^n + 3^n) \space \space \text{(for even $n$)}$$

Your sum is only particularly difficult because of the fact that the index begins negative, and the exponent begins at zero. Hopefully these general principles give you a feel for what is happening a bit better.