# Thread: Ratios of Binomial Expansions

1. ## Ratios of Binomial Expansions

1) In the expansion of (1+x)^16, fing the ratio of the term in x^13 to the term in x^11.

2) In the expansion of (2+3x)^n, the coefficients of x^5 and x^6 are in the ratio 4:9. Find the value of n.

Any help would be great, thanks!

2. ## Re: Ratios of Binomial Expansions

$(1+x)^{16} = \sum_{k=0}^{16} \binom{16}{k} 1^{16-k}x^k$

We want to find the terms in $x^{13}$ and $x^{11}$.

These are:

$\binom{16}{13}x^{13} \quad \mathrm{and} \quad \binom{16}{11}x^{11}.$

Ratio:

$\frac{\binom{16}{13}}{\binom{16}{11}} = \frac{\frac{16!}{3! \cdot 13!}}{\frac{16!}{5! \cdot 11!}} = \frac{5! \cdot 11!}{3! \cdot 13!} = \frac{5 \cdot 4}{12 \cdot 13} = \frac{5}{39}$

Note that the ratio should be less than 1, because for $(1+x)$ raised to any power, the coefficients will be largest in the "middle" of the expansion.

The expansion for 2) will be

$(2+3x)^{x} = \sum_{k=0}^{n} \binom{n}{k} 2^{n-k}(3x)^k = \sum_{k=0}^{n}\left( \binom{n}{k} 2^{n-k} 3^k\right)x^k$

You are given that the ratio of the co-efficients of $x^5$ and $x^6$ is $4/9$.

Can you take it from here?

3. ## Re: Ratios of Binomial Expansions

Originally Posted by fan96
$(1+x)^{16} = \sum_{k=0}^{16} \binom{16}{k} 1^{16-k}x^k$

We want to find the terms in $x^{13}$ and $x^{11}$.

These are:

$\binom{16}{13}x^{13} \quad \mathrm{and} \quad \binom{16}{11}x^{11}.$

Ratio:

$\frac{\binom{16}{13}}{\binom{16}{11}} = \frac{\frac{16!}{3! \cdot 13!}}{\frac{16!}{5! \cdot 11!}} = \frac{5! \cdot 11!}{3! \cdot 13!} = \frac{5 \cdot 4}{12 \cdot 13} = \frac{5}{39}$

Note that the ratio should be less than 1, because for $(1+x)$ raised to any power, the coefficients will be largest in the "middle" of the expansion.

The expansion for 2) will be

$(2+3x)^{x} = \sum_{k=0}^{n} \binom{n}{k} 2^{n-k}(3x)^k = \sum_{k=0}^{n}\left( \binom{n}{k} 2^{n-k} 3^k\right)x^k$

You are given that the ratio of the co-efficients of $x^5$ and $x^6$ is $4/9$.

Can you take it from here?
I should be able to;
When you simplify 16C13/16C11 and change it into factorial notation - how does the 5! and 11! get on the denominator while 3! and 13! get on the the denominator? Sorry if its a dumb question - just need an explanation please.

4. ## Re: Ratios of Binomial Expansions

$\frac{\frac{1}{5! \cdot 11!}}{\frac{1}{3! \cdot 13!}} = \frac{1}{5! \cdot 11!} \div \frac{1}{3! \cdot 13!} = \frac{1}{5! \cdot 11!} \times {3! \cdot 13!}$

Sometimes if you have nested fractions they can be very confusing.

Another way to think about it is to multiply both the numerator and denominator by $5! \cdot 11! \cdot 3! \cdot 13!$.

5. ## Re: Ratios of Binomial Expansions

Originally Posted by fan96
$\frac{\frac{1}{5! \cdot 11!}}{\frac{1}{3! \cdot 13!}} = \frac{1}{5! \cdot 11!} \div \frac{1}{3! \cdot 13!} = \frac{1}{5! \cdot 11!} \times {3! \cdot 13!}$

Sometimes if you have nested fractions they can be very confusing.

Another way to think about it is to multiply both the numerator and denominator by $5! \cdot 11! \cdot 3! \cdot 13!$.
Yeah - Thank you so much.

6. ## Re: Ratios of Binomial Expansions

Even though fan96 ninjad me lol

7. ## Re: Ratios of Binomial Expansions

Originally Posted by HeroWise
Even though fan96 ninjad me lol

Thank you so much

8. ## Re: Ratios of Binomial Expansions

Originally Posted by HeroWise
Even though fan96 ninjad me lol

Hey HeroWise, the answer is apparently 14 for this one - idk why - the working out looks fine to me (The question is from 5D in Cambridge 3U Year 12)

9. ## Re: Ratios of Binomial Expansions

I redid it im still gettign 9 i think text book errata

10. ## Re: Ratios of Binomial Expansions

Originally Posted by HeroWise
Even though fan96 ninjad me lol

Originally Posted by Heresy
Hey HeroWise, the answer is apparently 14 for this one - idk why - the working out looks fine to me (The question is from 5D in Cambridge 3U Year 12)
Originally Posted by HeroWise
I redid it im still gettign 9 i think text book errata
$\noindent The error was after your line T_{6}:T_{7} = 4:9, you forgot to include the ratio \frac{4}{9} when equating the expressions. That is, you wrote \color{red}\binom{n}{5}2^{n-5}3^{5} = \binom{n}{6}2^{n-6}3^{6}, but should have written \binom{n}{5}2^{n-5}3^{5} = \color{blue}\frac{4}{9}\times\color{black}\binom{n }{6}2^{n-6}3^{6}.$

11. ## Re: Ratios of Binomial Expansions

$\noindent By the way, to check your answer, you don't need to redo the whole question. You can just check by substituting your answer of \color{red}n=9 into your formulas for T_6 and T_7 and seeing whether \frac{T_6}{T_7} equals 4/9. This is a much faster way to check the answer, which could potentially help save time in exams.$

12. ## Re: Ratios of Binomial Expansions

oml I just realised, what i did there lol ill fix it asap. Was probs high while doing it lolll

13. ## Re: Ratios of Binomial Expansions

Originally Posted by InteGrand
$\noindent By the way, to check your answer, you don't need to redo the whole question. You can just check by substituting your answer of \color{red}n=9 into your formulas for T_6 and T_7 and seeing whether \frac{T_6}{T_7} equals 4/9. This is a much faster way to check the answer, which could potentially help save time in exams.$
Thanks for the tip!

14. ## Re: Ratios of Binomial Expansions

Originally Posted by HeroWise
oml I just realised, what i did there lol ill fix it asap. Was probs high while doing it lolll
Haha. Thanks for helping anyways - really appreciate it!

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