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I am an undergraduate and recently tried to find an expression to calculate the term $(y+x)^n-(y-x)^n$ for any $n \in \mathbb{N}$. I ended up with an expression for every even $n$ which is $$ (y+x)^{2n}-(y-x)^{2n}=2 \sum_{k=1}^n{2n \choose 2k-1} x^{2n-2k+1}\space y^{2k-1} $$ For the proof of this I started to split up the result of the binomial theorem for all even integers $n$ into a sum of even and uneven powers. I wanted to show that

$$ \sum^{2n}_{k=0}{{2n \choose k}x^{2n-k} \space y^{k}}= \sum_{k=0}^n{2n \choose 2k} x^{2n-2k}\space y^{2k} + \sum_{k=1}^n{2n \choose 2k-1} x^{2n-2k+1}\space y^{2k-1} \quad \forall \space n \in \mathbb{N}, n\geq1 $$ with mathematical induction.

The idea was to show that the term $\sum_{k=0}^n{2n \choose 2k} x^{2n-2k}\space y^{2k}$ is equal for $(y+x)^n$ aswell as $(y-x)^n$

While it is easy to show that the equation holds for $n=1$ I couldn´t finish the induction step.

With $ n \to n+1 $ we have

$$ \sum^{2(n+1)}_{k=0}{{2(n+1) \choose k}x^{2(n+1)-k} \space y^{k}}=\sum^{2n+2}_{k=0}{{2n+2 \choose k}x^{2n+2-k} \space y^{k}} $$

To use the induction hypothesis I tried to form that sum into $$ \sum^{2n}_{k=0}{{2n \choose k}x^{2n-k} \space y^{k}} $$ When doing so I fail to convert the binomial coefficient ${2n+2 \choose k}$ into a form of ${2n \choose k}+z$

Where z is the additional factor. In order to factor $z$ out of the sum it cannot be dependent on $k$.

Q: How can one modify the sum to get the desired result in order to use the induction hypothesis? Is there a better way to proof this result or is it just wrong?

I tried to factor it out directly and also tried to shift the sum but both cases yield a $z$ dependent on $k$

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  • $\begingroup$ Are you sure you need to prove that identity by induction? You're just expressing $\{0,1,2,\dots,2n\}=\{0,2,4,\dots,2n\}\cup\{1,3,5,\dots,2n-1\}$. $\endgroup$ – egreg Aug 5 '16 at 20:39
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Use the usual binomial theorem: $(a+b)^p=\sum_{j=0}^p\binom{p}{j}b^{n-j}a^j$:

With $a=y$, $b=x$, $p=2n$, we get $$(x+y)^{2n}=\sum_{j=0}^{2n}\binom{2n}{j}x^{2n-j}y^j$$ and with $a=y$, $b=-x$ and $p=2n$, we get $$(x-y)^{2n}=\sum_{j=0}^{2n}\binom{2n}{j}(-1)^jx^{2n-j}y^j$$ Now subtract the second equation from the first one, and the terms with $j$ even will cancel out, that is, you will end with a summation from $j=1,3,\ldots,2n-1$. Substitute $j=2k-1$ and you obtain the result.

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  • $\begingroup$ I appreciate the help here. Its way easier that one - this answers one part of my question. In case i would try to change the binomial coefficient the way i wanted too do you have any suggestion? $\endgroup$ – K. Hoffmann Aug 5 '16 at 21:05

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