If the co-efficients of \(x^{7}\) and \(x^{8}\) in the expansion of \(\left(2+\frac{x}{3}\right)^{n}\) are equal, then the value of \(n\) is equal to \_\_\_\_

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Binomial Theorem

If the co-efficients of

x^{7}

and

x^{8}

in the expansion of

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

are equal, then the value of

n

is equal to ____

Answer: 55

The general term in the expansion is

T_{r + 1} = (r n) 2^{n - r} (\frac{x}{3})^{r} = (r n) 2^{n - r} \frac{x ^{r}}{3 ^{r}}

. The coefficient of

x^{7}

(7 n) 2^{n - 7} \frac{1}{3 ^{7}}

and the coefficient of

x^{8}

(8 n) 2^{n - 8} \frac{1}{3 ^{8}}

. Setting them equal:

(7 n) 2^{n - 7} \frac{1}{3 ^{7}} = (8 n) 2^{n - 8} \frac{1}{3 ^{8}}

. Simplifying:

(7 n) \cdot 2 \cdot 3 = (8 n)

, i.e.,

6 (7 n) = (8 n)

. Using

(8 n) = \frac{n - 7}{8} (7 n)

, we get

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

, so

48 = n - 7

, giving

n = 55

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Numerical answer