Find $'a'$ if the $17^{ ext{th}}$ and $18^{ ext{th}}$ terms of the expansion $(x + a)^{50}$ are equal.

Question

Find $'a'$ if the $17^{	ext{th}}$ and $18^{	ext{th}}$ terms of the expansion $(x + a)^{50}$ are equal.

Accepted Answer

>

> [!TLDR]
> $a=\dfrac x2$.
>
> The general term of $(x+a)^{50}$ is $T_{r+1}=\binom{50}{r}x^{50-r}a^r$.
>
> So $T_{17}=\binom{50}{16}x^{34}a^{16}$ and $T_{18}=\binom{50}{17}x^{33}a^{17}$.
>
> Setting them equal:
> $$\binom{50}{16}x^{34}a^{16} = \binom{50}{17}x^{33}a^{17}$$
> $$\binom{50}{16}x = \binom{50}{17}a$$
> $$a = x\cdot\dfrac{\binom{50}{16}}{\binom{50}{17}}.$$
> …

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.Find $'a'$ if the $17^{\text{th}}$ and $18^{\text{th}}$ terms of the expansion $(x + a)^{50}$ are equal.

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 $y$ ? …

Q.Find ′a′'a'′a′ if the 17th17^{\text{th}}17th and 18th18^{\text{th}}18th terms of the expansion (x+a)50(x + a)^{50}(x+a)50 are equal.

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Q.Find $'a'$ if the $17^{\text{th}}$ and $18^{\text{th}}$ terms of the expansion $(x + a)^{50}$ are equal.

Unlock everything free for 14 days