What is the value of the finite sum?

Question

What is the value of the following sum:

$$1-3+5-...+1993-1995+1997?$$

I realize that regrouping the series' first and last terms, second and penultimate terms and so on gives $1998$ and $-1998$. I'm stuck there, however, and I don't know how to determine how much of the sum cancels itself out in the middle part of the sum.

I have tried looking at simpler sums, such as $1-3+5-7+9-11+13$, which obviously equals $7$, but I don't know how to apply that to my larger sum.

Lastly, is there a general formula for the alternating sum of the first n odd numbers?

$$\sum_{k=1}^{n} (-1)^{k+1}(2k-1)$$

Answer 1

You could also separate is into two part, the positive values and the negative values:

$$(1+5+...1993+1997)-(3+7+...1991+1995)$$ The first part has $\frac{1997-1}{4}=499$ terms and the second part has $\frac{1995-3}{4}=498$ terms. Also notice if we now pair the first and last term we get 1998, the second and second last term we get 1998, etc. So there are $\frac{499}{2}$ pairs in the first group that add up to 1998 and $\frac{498}{2}$ pairs in the second group that add up to 1998. We have $$=\frac{499}{2}\times 1998-\frac{498}{2}\times 1998$$ $$=1998\left(\frac{499}{2}-\frac{498}{2}\right)$$ $$=1998\times \frac{1}{2}$$ $$=999$$

Answer 2

Pair the terms and put brackets around each pair $$(1-3)+(5-7)+(9-11)+\cdots +1997$$ Each term inside the brackets evaluates to $-2$. The final term with positive sign is unpaired. Assuming the number of pairs is $n$ the sum is $-2n+1997$. In case there are are no dangling term at the right it will be simply $-2n$.

Answer 3

Selected answer is my favourite approach to the problem. For what it's worth, here's a nice way to write it in Sigma notation, though:

$$\sum_{i=0}^{499}(4i+1)-\sum_{i=0}^{498}(4i+3)$$

Answer 4

$$\begin{align} &\;\;\;\;\overbrace{1\;-3+5\;\;-7+9\;\;-11+13-\cdots \;\;-1995+1997}^{999\;\text{terms}}\\\\ =&\;\;\;\;1\;-3+5\;\;-7+9\;\;-11+13-\cdots \;\;-1995+1997\\ &\;\;\color{red}{+\underbrace{0+(4-4)+(8-8)+(12-12)-\cdots+(1996-1996)}_0} \\ =&\;\;\;\;\overbrace{1\;+1\;\;+1\;+1\;+1\;\;+1\;\;+1+\cdots \qquad\;+1\quad\;+1}^{999}\\\\ =&\;\;\;999\quad\blacksquare\end{align}$$

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From the above it can be seen that, for the general case of the alternating odd number series, if the first term is positive and there is an odd number of terms, the sum is equal to the number of terms.

Answer 5

It's simply

$$\frac{1}{2}\ (1997 + 1) = 999$$