Ratios of Binomial Expansions (1 Viewer)

Heresy · Nov 3, 2018

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!

fan96 · Nov 3, 2018

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

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?

Heresy · Nov 3, 2018

fan96 said:
$(1+x)^{16} = \sum_{k=0}^{16} \binom{16}{k} 1^{16-k}x^k$

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.

fan96 · Nov 3, 2018

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

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 $5! \cdot 11! \cdot 3! \cdot 13!$ .

Heresy · Nov 3, 2018

fan96 said:
$\frac{\frac{1}{5! \cdot 11!}}{\frac{1}{3! \cdot 13!}} = \frac{1}{5! \cdot 11!} \div \frac{1}{3! \cdot 13!} = \frac{1}{5! \cdot 11!} \times {3! \cdot 13!}$

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.

HeroWise · Nov 3, 2018

Even though fan96 ninjad me lol

Heresy · Nov 4, 2018

HeroWise said:
Even though fan96 ninjad me lol

Thank you so much

Heresy · Nov 4, 2018

HeroWise said:
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)

HeroWise · Nov 4, 2018

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

InteGrand · Nov 4, 2018

HeroWise said:
Even though fan96 ninjad me lol

Heresy said:
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)

HeroWise said:
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 $\binom{n}{5}2^{n-5}3^{5} = \color{blue}\frac{4}{9}\times\color{black}\binom{n}{6}2^{n-6}3^{6}$.$

InteGrand · Nov 4, 2018

HeroWise · Nov 4, 2018

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

Heresy · Nov 4, 2018

InteGrand said:
$\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!

Heresy · Nov 4, 2018

HeroWise said:
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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Ratios of Binomial Expansions (1 Viewer)

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