Finding $n>2$ such that $a_1$, $a_2$, $a_3$ are in arithmetico-geometric progression and $\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r}$

Question

This is the strangest, mind-boggling question I came across while doing binomial theorem.

Let $$\displaystyle\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r},\quad n\gt2$$
then find the value if $n$ so that $a_1,a_2,a_3$ are in A. G. P.

(original image)

Sorry but i have absolutely no clue of solving it.
max i did was to expand the stuff inside the summation but it made things worse by bringing in two summations. I had aimed at comparing coefficients.
can someone help me out ? hints appreciated.

** AGP = arithmetico geometric progression
** problem encountered in my JEE coaching study material

Answer 1

Dividing both sides by $(1-x)^{3n}$ shows that the equation is equivalent to $$ \left(1+\frac{3x}{(1-x)^2}\right)^n=\sum_{r=0}^na_r\left(\frac{x}{(1-x)^2}\right)^r\tag1 $$ Set $u=\frac{x}{(1-x)^2}$, then we have $$ \left(1+3u\right)^n=\sum_{r=0}^na_ru^r\tag2 $$ So $a_r=3^r\binom{n}{r}$, as ZAhmed computed.

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Suppose that $3\binom{n}{1},9\binom{n}{2},27\binom{n}{3}$ are in anArithmetico-Geometric Progression $$ a_0,a_0(1+\Delta)\mathrm{R},a_0(1+2\Delta)R^2\tag3 $$ Then $$ a_0=3n\tag4 $$ and $$ \frac{1+2\Delta+\Delta^2}{1+2\Delta}=\frac{\binom{n}{2}^2}{\binom{n}{1}\binom{n}{3}}=\frac32\frac{n-1}{n-2}\tag5 $$ Let $\lambda=\frac32\frac{n-1}{n-2}-1$ and then $$ \Delta=\lambda+\sqrt{\lambda(\lambda+1)}\tag6 $$ The ratio of the first two terms is $$ (1+\Delta)\mathrm{R}=\frac{9\binom{n}{2}}{3\binom{n}{1}}=\frac32(n-1)\tag7 $$ Therefore, $$ R=\frac{3(n-1)}{2(1+\Delta)}\tag8 $$

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Examples

These can be fit to an AGP for any $n\gt2$:

For $n=3$, we get $$ a_0=9,\Delta=2+\sqrt6,\mathrm{R}=\frac3{3+\sqrt6}\tag{Ex3} $$ For $n=4$, we get $$ a_0=12,\Delta=\frac{5+3\sqrt5}4,R=\frac6{3+\sqrt5}\tag{Ex4} $$ For $n=5$, we get $$ a_0=15,\Delta=1+\sqrt2,R=\frac6{2+\sqrt2}\tag{Ex5} $$ For $n=6$, we get $$ a_0=18,\Delta=\frac{7+\sqrt{105}}8,R=\frac{60}{15+\sqrt{105}}\tag{Ex6} $$ For $n=7$, we get $$ a_0=21,\Delta=2,R=3\tag{Ex7} $$

Answer 2

Let us find out $A_r$ by rearranging both the sides as $$(1-x^3)^n=\sum_{r=0}^{n} A_r x^r(1-x)^{3n-3r} \implies\left(1+3\left(\frac{x}{1-x)^2}\right)\right)^n =\sum_{r=0}^{n}A_r \left(\frac{x}{(1-x)^2}\right)^r $$ $$\implies (1+3y)^n=\sum_{r-0}^{n} A_r y^r \implies \sum_{r=0}^{n} {n \choose r} 3^r y^r=\sum_{r=0}^{n} A_r y^r \implies A_r=3^r{n \choose r}$$ If $A_1,A_2,A_3$ form an AGP, then ${n \choose 1}, {n \choose 2}, {n\choose3}$ should be in AP $$\implies 6n(n-1)=6n+ n(n^2-3n+2) \implies n^3-9n+14 \implies n=2,7$$ $n=7$ is acceptable.

Answer 3

We'll start with writing $(1-x^3)^n$ as a sum :$$(1-x^3)^n = \sum_{r=0}^{n} (-1)^r (x)^{3r}=\sum_{r=0}^{n} a_r x^r\frac{(1-x)^{3n}}{(1-x)^{2r}}$$ Now both have equal terms. So we can actually the equate the corresponding terms. So, $$(-1)^r x^{3r} = a_r x^r \frac{(1-x)^{3n}}{(1-x)^{2r}}$$ Now transposing the $x^r$ term to the LHS, we get : $$(-1)^r x^{2r} = a_r \frac{(1-x)^{3n}}{(1-x)^{2r}}$$ Now, transpose the $x^{2r}$ to the RHS, we get : $$(-1)^r = a_r \frac{(1-x)^{3n}}{x^{2r}(1-x)^{2r}}$$ $$= a_r \frac{(1-x)^{3n}}{(x^2-x)^{2r}}$$ $$\implies a_r = (-1)^r \frac{(x^2-x)^{2r}}{(1-x)^{3n}}$$ And that is the value of $a_r$. Now I guess, you can take it from here

Hope it helps