How do I convert $-1 + 3 - 5 +...- 101$ into sigma notation.

I tried to divide the series into $-1 -5 -7 -...$ and $3+7+9$ but i'm not too sure if that is correct.

  • 1
    $\begingroup$ Think that they are odd numbers and about $(-1)^n$. $\endgroup$ – Claude Leibovici Oct 6 '17 at 12:36


  • $\begingroup$ Yes that is the answer, but i don't get where it comes from? $\endgroup$ – Sumeet Singh Oct 6 '17 at 12:38
  • $\begingroup$ @Sumeet Singh $(-1)^k$ gives changing of signs. $2k-1$ it's $a_k$ in the arithmetic progression. $a_1=1$ and $d=2$. $a_k=a_1+(k-1)d.$ $\endgroup$ – Michael Rozenberg Oct 6 '17 at 12:40
  • $\begingroup$ Oh that makes sense now. Thank you very much $\endgroup$ – Sumeet Singh Oct 6 '17 at 12:43
  • $\begingroup$ @Sumeet Singh You are welcome! $\endgroup$ – Michael Rozenberg Oct 6 '17 at 12:53

First, take care of the alternating signs. What you need is some function that changes sign for each integer, so something that maps $1$ to $-1$, $2$ to $1$, $3$ to $-1$ and so on. Luckilly, that's easy, since it's just $(-1)^k$. So that gives you

$$\begin{align}-1+3-5+7-\cdots -101 &= (-1)^1\cdot 1 + (-1)^2\cdot 3 +(-1)^3\cdot 5+\cdots (-1)^{51}\cdot 101 \end{align}$$

Now you need something that maps:

$$\begin{align}1&\mapsto 1\\ 2&\mapsto 3\\ 3&\mapsto 5\\ 4&\mapsto 7\\ &\vdots\\ 51&\mapsto 101\end{align}$$

and you can quickly see that that would be $k\mapsto (2k-1)$, so that gives you

$$(-1)^1\cdot 1 + (-1)^2\cdot 3 +(-1)^3\cdot 5+\cdots (-1)^{51}\cdot 101 \\= (-1)^1\cdot (2\cdot 1 -1) + (-1)^2\cdot(2\cdot 2-1) + (-1)^3+(2\cdot 3-1)+\cdots + (-1)^{51}\cdot(2\cdot 51-1)$$

and this can more simply be written as $$\sum_{k=1}^{51} (-1)^k(2\cdot k-1).$$

  • $\begingroup$ Thank you! It was really helpful. $\endgroup$ – Sumeet Singh Oct 6 '17 at 12:54
  • $\begingroup$ @SumeetSingh Happy to help. $\endgroup$ – 5xum Oct 6 '17 at 12:54

Look at each term individually and think about how to express the term as a function of its position in the sum. Try to look for a pattern that can be applied to all terms.

The first term can be considered position #1 or position #0. Mostly it's personal preference. Sometimes one way is better (i.e., simpler) than the other. Sometimes you're told to do it one way or another. Let's call the first term position #1 here.

Here's a chart of the first few terms with their corresponding position numbers.

\begin{array}{c|c|c|c} \text{position } (k) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{term} & -1 & 3 & -5 & 7 & -9 & 11 \end{array}

First we notice that the terms are alternating in sign. The first term is negative, the next is positive, the next is negative, and so on. This tells us that our general formula will have $(-1)^k$ or $(-1)^{k+1}$ somewhere in it, because that's how we obtain alternating signs in general. Our terms are all negative when $k$ is odd. They're all positive when $k$ is even. $(-1)^k$ matches this pattern. Therefore we need $(-1)^k$ in our expression.

That takes care of the signs. Now we can ignore them. In other words, we can continue to search for a general pattern by pretending all the terms are positive. Here's the above chart with no minus signs:

\begin{array}{c|c|c|c} \text{position } (k) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{|term|} & 1 & 3 & 5 & 7 & 9 & 11 \end{array}

I put absolute value symbols around "term" just to be thorough and accurate. So now we ask ourselves, for each term, what do we need to do to $k$ to get the value of |term|? What do we need to do to $k=1$ to get $1$? What do we need to do to $k=2$ to get $3$? What do we need to do to $k=3$ to get $5$? And so on. And our answer needs to be the same for every term. There are a lot of things we can do to $k=1$ to get $1$. And a lot of things we can do to $k=2$ to get $3$, etc. But there is (ideally) only one thing we can that works for all of our values of $k$ that we need. This part is basically pattern identification. It requires some mathematical skill but it gets easier with practice. The pattern I see (and has already been pointed out elsewhere here) is that we can get the value of |term| by doubling $k$ and then subtracting $1$. When $k = 1$, we see that $2k-1 = 1$. When $k = 2$, we have $2k-1 = 3$. When $k=3$, we have $2k-1 = 5$. And so on. Looks like we've found our general expression!

Combining the two results above, our general expression is $(-1)^k(2k-1)$. Finally, where do we start and where do we stop the summation? We start at $k=1$ because we called our first term position #1, i.e., the position where $k=1$. We stop when our term is $-101$. Which $k$ does this correspond to? Just set $2k-1 = 101$ and solve for $k$, and we get $k=51$. So, overall, our summation is: $$ \sum_{k=1}^{51} (-1)^k (2k-1)$$


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