4
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ Welcome to MSE. Please type your questions rather than posting images. Images can't be browsed, and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial To begin with, surround all math expressions (including numbers,) with $ signs. Use ^ for exponents and _ for subscripts. $x_1^{2/3}$ shows up as $x_1^{2/3}$. $\endgroup$
    – saulspatz
    Jun 7, 2021 at 15:33
  • 2
    $\begingroup$ It would be nice to include where you encountered this question and, in particular, what A. G. P. means. $\endgroup$
    – robjohn
    Jun 7, 2021 at 15:38
  • $\begingroup$ Comparing coefficients sounds right, but I don't know that you have to multiply it all out. For example, the leading coefficient on the left -hand side is $(-1)^n$ and the leading coefficient on the right-hand side is $a_0(-1)^{3n}$ so $a_0=1$. See if you can work out the next few in a similar way. $\endgroup$
    – saulspatz
    Jun 7, 2021 at 15:49
  • $\begingroup$ If I'm not mistaken, any three numbers can be fit to an AGP. $\endgroup$
    – robjohn
    Jun 7, 2021 at 16:57

3 Answers 3

2
$\begingroup$

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.


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, $$ R=\frac{3(n-1)}{2(1+\Delta)}\tag8 $$


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 $$ a_0=21,\Delta=2,R=3\tag{Ex7} $$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Yes, I have edited my answer one should get $n=7$. $\endgroup$
    – Z Ahmed
    Jun 7, 2021 at 17:08
  • 1
    $\begingroup$ technically as robjohn points, we can have infinite n values giving infinite AGP's but since it was JEE integer type question, 7 is correct ! $\endgroup$ Jun 8, 2021 at 3:06
0
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ $a_r$ cannot depend ob $x$. $\endgroup$
    – Z Ahmed
    Jun 7, 2021 at 17:19
  • $\begingroup$ yes, algebrology, please note that Ar doent depend on x. $\endgroup$ Jun 8, 2021 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.