If the co-efficient of x7 and x8 in the expansion of ( 2 + x \over 3 )^n are equal, then the value of n is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q46 with Solution

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Correct answer: 55

For

\left(2+\frac{x}{3}\right)^n,

the general term is

T_{r+1}=\binom{n}{r} 2^{\,n-r}\left(\frac{x}{3}\right)^r.

So the coefficient of $x^r$ is

\binom{n}{r}2^{\,n-r}\frac{1}{3^r}.

Coefficient of $x^7$ :

\binom{n}{7}2^{\,n-7}\frac{1}{3^7}

Coefficient of $x^8$ :

\binom{n}{8}2^{\,n-8}\frac{1}{3^8}

Given these are equal,

\binom{n}{7}2^{n-7}\frac{1}{3^7} = \binom{n}{8}2^{n-8}\frac{1}{3^8}.

Multiply both sides by $3^8$ :

3\binom{n}{7}2^{n-7} = \binom{n}{8}2^{n-8}.

Now divide both sides by $2^{n-8}$ :

6\binom{n}{7} = \binom{n}{8}.

Using

\binom{n}{8} = \binom{n}{7}\cdot \frac{n-7}{8},

we get

6\binom{n}{7} = \binom{n}{7}\cdot \frac{n-7}{8}.

Assuming $\binom{n}{7} \neq 0$ , cancel it:

6 = \frac{n-7}{8}.

So,

n-7=48

n=55.

Therefore,

\boxed{55}