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Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$

My try:

I assumed $A=2x^2$,$B=3x$ and $C=4$

Then we have the following cases to collect coefficient of $x^7$:

Case $1.$ $A^3 \times B^1 \times C^6$

Case $2.$ $A^2 \times B^3 \times C^5$

Case $3.$ $A^1 \times B^5 \times C^4$

case $4.$ $A^0 \times B^7 \times C^3$

Using multinomial theorem we get Coefficient of $x^7$ as:

$$a_7=10! \times \left(\frac{2^33^14^6}{3!6!}+\frac{2^13^54^4}{1!5!4!}+\frac{2^23^34^5}{2!3!5!}+\frac{2^03^74^3}{3!7!}\right)$$

Like wise we need to find $a_{13}$

But is there any better way?

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  • $\begingroup$ Replace $$x\rightarrow \frac{1}{x}$$ $\endgroup$ – DXT Apr 28 at 19:14
  • $\begingroup$ @DXT, what is your approach after that? $\endgroup$ – i707107 Apr 28 at 19:22
  • $\begingroup$ Same as above done by Umesh $\endgroup$ – DXT Apr 28 at 19:34
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This approach is from the idea of Pascal's triangle. We write $$ (2x^2+3x+4)^n = \sum_r a_{n,r} x^r $$

Then $$ (2x^2+3x+4)^{n+1}=(\sum_r a_{n,r} x^r )(2x^2+3x+4) = \sum_r (2a_{n,r-2}+3a_{n,r-1}+4a_{n,r})x^r.$$ Thus, by comparing coefficients, we obtain a recurrence relation $$ a_{n+1,r}=2a_{n,r-2}+3a_{n,r-1}+4a_{n,r} $$ with an initial condition $a_{0,0}=1$, $a_{1,2}=2$, $a_{1,1}=3$, and $a_{1,0}=4$.

I am not sure if this is a better way. But, it will be more convenient than computing factorials many times. Especially for small numbers, this method will give answers quickly.

However, for large numbers, the number of recursion steps will increase enormously with $n$.

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Given $$(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}x^k$$

Now using $\displaystyle x\rightarrow \frac{2}{x}$

$$\Rightarrow 2^{10}(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k}x^{20-k}$$

$$\Rightarrow (2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k-10}x^{20-k}$$

Now Comparing Coefficient of $x^{7}$ on both side, we get

$$a_{7}=a_{13}\cdot 2^{3}\Rightarrow \frac{a_{7}}{a_{13}}=8.$$

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  • $\begingroup$ This is very nice that you didn't have to find all coefficients. $\endgroup$ – i707107 Apr 29 at 15:28
  • $\begingroup$ Excellent approach. Kudos. But i guess this approach not applicable when question is asked to find $\frac{a_5}{a_9}$ $\endgroup$ – Umesh shankar Apr 29 at 16:29

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