Find the coefficients of x^7 and x^8 in the expansion of (2+x/3)^n

Question

Find the coefficients of x⁷ and x⁸ in the expansion of \(\Big(2+\frac{x}{3}\Big)^n\)

Answer 1

To find : coefficients of x⁷ and x⁸ 

Formula :  t_r+1 = (_rⁿ)a^n-rb^r

Here, a=2, b = \(\frac{x}{3}\)

We have, t_r+1 = (_rⁿ)a^n-rb^r

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

= (_rⁿ) \(\frac{2^{n-r}}{3^r}\) x^r

To get a coefficient of x⁷ , we must have, 

x⁷ = x^r 

• r = 7 

Therefore, the coefficient of x⁷ = = (_rⁿ) \(\frac{2^{n-7}}{3^7}\)  

And to get the coefficient of x⁸ we must have, 

x⁸ = x^r 

• r = 8 

Therefore, the coefficient of x⁸ =  (₈ⁿ) \(\frac{2^{n-8}}{3^8}\)

Conclusion :

• Coefficient of x⁷ =  (₇ⁿ) \(\frac{2^{n-7}}{3^7}\)

• Coefficient of x⁸ = (₈ⁿ) \(\frac{2^{n-8}}{3^8}\)