Write 3 - 5 + 7 - 9 + 11 - 13 + 15 in sigma notation.

Question

parthkohli · Answer

$$\textbf{HINT}$$You'd have to use the powers of $-1$ because there are sign changes in your series. Also, use the property of odd numbers.

anonymous · Answer

Meaning....? Sorry I'm really lost here

parthkohli · Answer

You know that every odd number is in the form $2n+1$ where $n$ is some integer.

anonymous · Answer

so the bottom would be n-1?

anonymous · Answer

$$\sum_{n-1}^{?}$$ right?

skullpatrol · Answer

No.

parthkohli · Answer

I'm not really sure how I'd explain this to you, so here's two exemplary hints.$$\sum_{\Large{n =0}}^{\Large 5}(-1)^n = (-1)^0 + (-1)^1 + (-1)^2\cdots(-1)^5 = 1 +(-1) + 1 \cdots(-1)$$and$$\sum_{\Large{n = 1}}^{\Large  5} 2n + 1 = 2(1) + 1 + 2(2) + 1\cdots2(5) + 1 = 3 + 5 + \cdots+11$$

parthkohli · Answer

Another (and probably the last) hint: $(-1)^1\times 5 = -5$ and $(-1)^2 \times 5 = 5$.

parthkohli · Answer

Now can you think about it a little? I'm ready to support you through the rest.

anonymous · Answer

I just found where we weren't suppose to do that problem. Sorry all!

parthkohli · Answer

lol okay

parthkohli · Answer

You're welcome BTW. =)

anonymous · Answer

Thanks for trying tho!