Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal

Question

Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.

My attempt:

$\displaystyle \binom{n}{7}=\binom{n}{8} $

$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$

$$ \frac{6}{7!} = \frac{n-7}{40320} $$

$$ n-7 = 48 $$

$$ n=55 $$

If $$\binom{n}{7}=\binom{n}{8} $$ then $n=7+8=15$ – Uri Goren Mar 06 '16 at 19:18 — Uri Goren, Mar 06 '16 at 19:18
Do any $n$ between $0$ and $6$ count? – Cameron Williams Mar 06 '16 at 19:34 — Cameron Williams, Mar 06 '16 at 19:34

score 3 · Accepted Answer · answered Mar 06 '16 at 19:21

3

The coefficient of $x^7$ is $$\binom{n}{7}\frac{2^{n-7}}{3^7}$$ And the coefficient of $x^8$ is $$\binom{n}{8}\frac{2^{n-8}}{3^8}$$ Comparing them we get: $$\binom{n}{8}=\binom{n}{7}\frac{3}{2}$$

answered Mar 06 '16 at 19:21

Uri Goren

2,940

score 1 · Answer 2 · answered Mar 06 '16 at 19:20

1

The general term of $(a+b)^n$ $$t_{r+1}={n\choose r}.a^r.b^{n-r}$$ plug in r as $6,7$ and you will get it

answered Mar 06 '16 at 19:20

Archis Welankar

15,910

score 1 · Answer 3 · answered Mar 06 '16 at 19:26

The coefficient of $x^7$ will be ${n\choose 7}{2^{n-7}\over 3^7}$ and the coefficient of $x^8$ will be ${n\choose 8}{2^{n-8}\over 3^8}$ and not what you have got .you just missed out the factor of 3 that plays a role as a coefficient.So equating it you get $${n\choose 7}{2^{n-7}\over 3^7}={n\choose 8}{2^{n-8}\over 3^8}$$ and get the answer as $55$

score 1 · Answer 4 · answered Mar 06 '16 at 19:31

Reference that $$(a + b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b^1 + \cdots + {n \choose n-1}ab^{n-1} + {n \choose n}a^0b^n.$$

So we want $a = 2$ and $b = {x \over 3}$. So we are considering the terms $\displaystyle {n \choose 7}a^{n - 7}b^7$ and $\displaystyle {n\choose 8}a^{n-8}b^8.$ So, $${n \choose 7}a^{n - 7}b^7 = {n\choose 8}a^{n-8}b^8$$ $${n! \over 7!(n - 7)!}a^{n-7}b^7 = {n! \over 8!(n-8)!}a^{n-8}b^8$$ $${a^{n-8}a \over 7!(n-7)(n-8)!} = {a^{n-8}b \over 8\times 7!(n-8)!}$$ $${a \over n-7} = {b \over 8}$$ $$n-7 = {8a \over b}$$ $$n = {48 \over x} + 7$$ Can you see where it comes in now?

score 1 · Answer 5 · answered Mar 06 '16 at 19:33

$\displaystyle \left(2+\frac{x}{3}\right)^{n}=\sum_{i=0} ^n {n \choose k}({\frac{x}{3}})^i 2^{n-i}$
Now for $i =7\implies{n \choose 7}({\frac{x}{3}})^7 2^{n-7}$

and for $i =8\implies{n \choose 8}({\frac{x}{3}})^8 2^{n-8}$

now to get the coefficients of $x^8$ & $x^7$ equal. $\implies$

${n \choose 7}({\frac{1}{3}})^7 2^{n-7}={n \choose 8}({\frac{1}{3}})^8 2^{n-8} \implies \frac{1}{n-7}=\frac{1}{8}* \frac{1}{3}*\frac{1}{2} \implies n=55$

Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal

5 Answers5