If coefficients of {x^7} and {x^8} in {left( {2 + frac{x}{3}} right)^n} are equal, then n is

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7.Binomial Theorem

easy

If the coefficients of ${x^7}$ and ${x^8}$ in ${\left( {2 + \frac{x}{3}} \right)^n}$ are equal, then $n$ is

A

$56$

B

$55$

C

$45$

D

$15$

Solution

(b) ${x^7}$,${x^8}$ will occur in ${T_8}$ and${T_9}$.

Coefficients of ${T_8}$ and ${T_9}$ are equal.

$\therefore {\,^n}{C_7}{2^{n – 7}}{\left( {\frac{1}{3}} \right)^7} = {\,^n}{C_8}{2^{n – 8}}{\left( {\frac{1}{3}} \right)^8}\,\, $

$\Rightarrow n = 55$.

Standard 11

Mathematics

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