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Show $${n \choose 1} + {n \choose 3} +\cdots = {n \choose 0} + {n \choose 2}+\cdots$$

A hint is given to consider the expansion $(x-y)^n$

However, when I plug in a number for $n$, I don't get an equality. $n=5$, for instance, I get $5+10 = 1 +10$.

How is this equality possible?

Michael Hardy
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user139181
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  • It might help if you write the equality properly. You're missing terms on both sides for $n=5$. – Git Gud Sep 11 '14 at 15:36
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    Since $n=5$ was mentioned as an alleged counterexample, let's look at it: $\displaystyle{5\choose0}+{5\choose 2}+{5\choose 4}$ $=1+10+5=16$ and $\displaystyle{5\choose1}+{5\choose3}+{5\choose5}=5+10+1=16$. The more interesting case is with an even number, since then you don't have the same list of terms being added: $\displaystyle{6\choose0}+{6\choose2}+{6\choose4}+{6\choose6}$ $=1+15+15+1=32$ and $\displaystyle{6\choose1}+{6\choose3}+{6\choose5}=6+20+6=32$. ${}\qquad{}$ – Michael Hardy Sep 11 '14 at 16:21
  • To see combinatorial viewpoint, see here: http://math.stackexchange.com/questions/248245/exactly-half-of-the-elements-of-mathcalpa-are-odd-sized – Martin Sleziak Sep 11 '14 at 17:20

6 Answers6

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Note that $$0=\{1+(-1)\}^n=\sum_{i=0}^{n}\binom{n}{i}\cdot 1^{n-i}\cdot (-1)^i.$$

mathlove
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Hint: $$\begin{pmatrix} n \\ k\end{pmatrix} = \begin{pmatrix} n \\ n-k \end{pmatrix}.$$

Emily
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The standard binomial expansion gives:

$$(x+y)^n=\sum_{r=0}^{n} \binom nr x^{n-r}y^r$$

Put $x=1, y=-1$: $$\begin{align} 0&=\sum_{r=0}^{n} \binom nr (-1)^r \\ &=\binom n0-\binom n1+\binom n2-\binom n3+\cdots +(-1)^n\binom nn\\ \binom n1+\binom n3+\binom n5+\cdots&=\binom n0+\binom n2+\binom n4\cdots\end{align}$$

It is interesting to consider some simple numerical examples of the expansion for different values of $n$ to get a more intuitive feel of the formula. A bar above the number indicates a negative sign.

$n=2$:

$$1\quad \bar{2}\quad 1$$

$n=3$:

$$1\quad \bar{3}\quad 3\quad \bar{1}$$

$n=4$:

$$1\quad \bar{4}\quad 6\quad \bar{4}\quad 1$$

$n=5$:

$$1\quad \bar{5}\quad 10\quad \overline{10}\quad 5\quad\bar{1}$$

From the above it is clear that:

  • for odd $n$, a coefficient has the opposite sign of its "mirror image", i.e. $${n\choose k}=-{n\choose n-k}$$ (e.g. 1, -1; -3, 3) thus cancelling out pairwise.

  • for even $n$, this does not occur as a coefficient has the same sign as its "mirror image" (e.g. 1, 1; -4, -4); however, the sum of coefficients in even positions is numerically equal to the sum of coefficients in odd positions, but is negative, thus the sums cancel out (e.g. 1+6+1=4+4, and (1+6+1)+(-4-4)=0)

Thus the formula holds for both odd and even $n$.

Hypergeometricx
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You know that:

$$(x+y)^n=\sum_{i=0}^{n}x^iy^{n-i} {n \choose i}$$

If you substitute $x=1$ and $y=-1$ you get:

$$0=(1-1)^n=\sum_{i=0}^{n}1^i \cdot (-1)^{n-i} {n \choose i}$$

Next move elements with odd index to left.

agha
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$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}=(1+(-1))^n=0$$

Adi Dani
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If you flip a fair coin $n\ge 1$ times. There are an equal probability, and hence an equal number of ways, of seeing an even number of heads as an odd number of heads. This is because after the first $n-1$ flips your heads parity is either even or odd, and on the $n^{th}$ flip there is a 50-50 chance of staying the same or changing.

paw88789
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