If the coefficients of 4 consecutive terms in the expansion of $(1 + x)^n$ are $a_1, a_2, a_3, a_4$ respectively, then show that $\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$.

Question

Accepted Answer

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> [!TLDR]
> Using $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$, each fraction $\dfrac{a_i}{a_i+a_{i+1}}$ simplifies to a linear expression, and LHS $=$ RHS.
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> Let the four consecutive coefficients be
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> $a_1=\binom{n}{r},\ a_2=\binom{n}{r+1},\ a_3=\binom{n}{r+2},\ a_4=\binom{n}{r+3}.$
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> Compute the building block:
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> $\dfrac{a_1}{a_1+a_2}=\dfrac{\binom{n}{r}}{\binom{n}{r}+\binom{n}{r+1}}=\dfrac{\binom{n}{r}}{\binom{n+1}{r+1}}.$
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> Now $\dfrac{\binom{n}{r}}{\binom{n+1}{r+1}}=\dfrac{n!/[r!(n-r)!]}{(n+1)!/[(r+1)!(n-r)!]}=\dfrac{r+1}{n+1}.$
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> By the same computation with $r	o r+1$ and $r	o r+2$:
> …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.If the coefficients of 4 consecutive terms in the expansion of $(1 + x)^n$ are $a_1, a_2, a_3, a_4$ respectively, then show that $\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$ .

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

What If It's Not Just $x$ and $y$ ? …

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