If the coefficients of $x^9$, $x^{10}$, $x^{11}$ in the expansion of $(1 + x)^n$ are in A.P., then prove that $n^2 - 41n + 398 = 0$.

Question

Accepted Answer

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> [!TLDR]
> Applying $2\,{}^{n}C_{10} = {}^{n}C_9 + {}^{n}C_{11}$ and using coefficient ratios yields $n^2 - 41n + 398 = 0$.
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> In $(1+x)^n$ the coefficients of $x^9, x^{10}, x^{11}$ are ${}^{n}C_9, {}^{n}C_{10}, {}^{n}C_{11}$. If they are in A.P.,
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> $2\,{}^{n}C_{10} = {}^{n}C_9 + {}^{n}C_{11}$.
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> Divide by ${}^{n}C_{10}$ and use $\dfrac{{}^{n}C_{r-1}}{{}^{n}C_r} = \dfrac{r}{n-r+1}$:
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> $\dfrac{{}^{n}C_9}{{}^{n}C_{10}} = \dfrac{10}{n-9}, \qquad \dfrac{{}^{n}C_{11}}{{}^{n}C_{10}} = \dfrac{n-10}{11}$.
>
> So
> …

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 $x^9$ , $x^{10}$ , $x^{11}$ in the expansion of $(1 + x)^n$ are in A.P., then prove that $n^2 - 41n + 398 = 0$ .

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.If the coefficients of x9x^9x9, x10x^{10}x10, x11x^{11}x11 in the expansion of (1+x)n(1 + x)^n(1+x)n are in A.P., then prove that n2−41n+398=0n^2 - 41n + 398 = 0n2−41n+398=0.

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Q.If the coefficients of $x^9$ , $x^{10}$ , $x^{11}$ in the expansion of $(1 + x)^n$ are in A.P., then prove that $n^2 - 41n + 398 = 0$ .

Unlock everything free for 14 days