Proof involving consecutive coefficients

Question

If $a_1,a_2,a_3,a_4$ are four consecutive coefficients in the expansion of $(1+x)^n$, prove that $$\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{2a_{2}}{a_{2}+a_{3}}$$

My solution:

Let $a_1 x^{r-1},a_2 x^{r},a_3 x^{r+1},a_4 x^{r+2} = {n\choose{r-1}} x^{r-1},{n\choose{r}} x^{r},{n\choose{r+1}} x^{r+1},{n\choose{r+2}} x^{r+2}$.

I sub this into the required proof and I have to do some laboured, long algebra process.

From my experience, someone on MSE will have a much more concise proof than this, so I am asking to enlighten myself.

Answer 1

This is just another variation where we stick at binomial coefficients avoiding factorials. We recall two binomial identities: \begin{align*} \binom{n+1}{r} &= \binom{n}{r-1}+\binom{n}{r}\tag{1}\\ \binom{n+1}{r}&=\binom{n}{r-1}\frac{n+1}{r}\tag{2} \end{align*}

Left-hand side: \begin{align*} \color{blue}{\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}} &=\frac{\binom{n}{r-1}}{\binom{n}{r-1} + \binom{n}{r}}+\frac{\binom{n}{r+1}}{\binom{n}{r+1} + \binom{n}{r+2}}\\ &=\frac{\binom{n}{r-1}}{\binom{n+1}{r}}+\frac{\binom{n}{r+1}}{\binom{n+1}{r+2}}\tag{$\to (1)$}\\ &=\frac{\binom{n}{r-1}}{\binom{n}{r-1}\frac{n+1}{r}}+\frac{\binom{n}{r+1}}{\binom{n}{r+1}\frac{n+1}{r+2}}\tag{$\to (2)$}\\ &=\frac{r}{n+1}+\frac{r+2}{n+1}\\ &\,\,\color{blue}{=\frac{2(r+1)}{n+1}} \end{align*} Right-hand side: \begin{align*} \color{blue}{\frac{2a_{2}}{a_{2}+a_{3}}}&=\frac{2\binom{n}{r}}{\binom{n}{r}+\binom{n}{r+1}} =\frac{2\binom{n}{r}}{\binom{n+1}{r+1}}\tag{$\to (1)$}\\ &=\frac{2\binom{n}{r}}{\binom{n}{r}\frac{n+1}{r+1}}\tag{$\to (2)$}\\ &\,\,\color{blue}{=\frac{2(r+1)}{n+1}} \end{align*} and the claim follows.

Answer 2

Hints:

Write the terms in the equality to be proved as follows:

$$\frac{a_k}{a_k+a_{k+1}}=\frac1{1+\cfrac{a_{k+1}}{a_k}}$$

Evaluate now each term separatedly. For example:

$$\frac{a_4}{a_3}=\frac{\binom n{r+2}}{\binom n{r+1}}=\frac{n!(r+1)! (n-r-1)!}{n!(r+2)!(n-r-2)!}=\frac{n-r-1}{r+2}\implies1+\frac{a_4}{a_3}=\frac{n+1}{r+2}\implies$$

$$ \frac1{1+\frac{a_4}{a_3}}=\frac{r+2}{n+1}$$

After you do this with the left side (and they have they same denominator), do the same with the right side...and it follows at once that they are the same

Answer 3

As Forester hinted, as you are interested in the coefficient, you should drop the $x$ monomials.

All that is needed here is basic operations on binomial coefficients: $$ \binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k} $$

In your case, you chose: $$ a_i = \binom{n}{r+i-2} $$

Then: $a_i + a_{i+1} = \binom{n+1}{r+i-1}$ and $\frac{a_i}{a_i + a_{i+1}} = \frac{r+i-1}{n+1}$.

Checking that the equation you're looking for holds is fairly straightforward: $$ \frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2r+2}{n+1} = 2\frac{a_2}{a_2 + a_3} $$

Answer 4

$$\frac{a_1}{a_1+a_2}=\frac{\binom{n}{r-1}}{\binom{n+1}{r}}=\frac{n!(n-r+1)!r!}{(r-1)!(n-r+1)!(n+1)!}=\frac{r}{n+1}$$

Similarly, $$\frac{a_3}{a_3+a_4}=\frac{r+2}{n+1}$$

So, summing gives:

$$\frac{2(r+1)}{n+1}$$

which equals $$\frac{2a_2}{a_2+a_3}$$