If the coefficients of `x^7 and x^8` in the expansion of `[2 +x/3]^n` are equal, then the value of n is : (A) 15 (B) 45 (C) 55 (D) 56

Answer

n=56

Explanation

We have , `T _(r+1) = .^(n) C _(r) 2 ^ ( n-r) xx (x/3)^(r) = .^(n)C_(r) xx 2 ^(n-r) xx 3 ^(-r) xx x^(r).`
Coeff. of `x^(7) = "coeff. of " x^(8) rArr .^(n) C _(7) xx 2^(n-7) xx 3^(-7) = .^(n) C _(8) xx 2^(n-8) xx 3^(-8)`
`:. (.^(n) C _(7))/(.^(n)C_(8)) = (2^(n-8) xx3^(-8))/(2^(n-7) xx 3^(-7))=1/(2 xx 3)= 1 / 6 rArr (n!)/((7!)(n-7)!) xx ((8!) xx ( n - 8)!)/(n!)=1/6.`
`:. 8/(n-8) = 1/6 rArr n-8 = 48 rArr n = 56.`