Test: Binomial Theorem: Simple Applications (May 30) - JEE MCQ

(x+a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^(n-1) a¹ + ⁿC₂ x^(n-2) a² + ..........+ ⁿC_(n-1) xa^(n-1) + ⁿC_n  aⁿ
Now, ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1,    ⁿC₂ = ⁿC_n-2,........
therefore, ⁿC_r = ⁿC_n-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

coefficient of x^-7 in [2+1/3x]ⁿ is ⁿC₇(2)^n-7 (1/3)⁷
coefficient of x^-8 in [2+1/3x]n is ⁿC₈(2)^n-8 (1/3)⁸
⟹ ⁿC₇ × (2^n-7) × 1/3⁷ = ⁿC₈ × (2^n-8) × 1/3⁸
⟹ ⁿC₈/ⁿC₇ = 6
⟹ (n-7)/8 = 6
⟹ n = 55

 Let Tr and Tr+1 denote the rthand(r+1)th terms in
the expansion of (1+x)¹⁹. 
 Tr = ¹⁹C_r-1 x^r-1 and T_r+1 = ¹⁹C_r x^r .
∴T_r+1/T_r = ¹⁹C_r xr/(19C_r-1 x^r-1)
⇒T_r+1/T_r = ¹⁹C_r  ¹⁹C_r-1 x
⇒T_r+1/T_r = 19!/(19−r)!r! × x[(19−r+1)!(r−1)]/10!
⇒T_r+1/Tr = x(20−r)/r
⇒T_r+1/T_r = (20−r)/r × 1/2   [∵x not equal to 1/2]
Now
T_r+1/T_r > 1
⇒ (20−r)/r × 1/2 > 1
⇒ 20 > 3r
r > 20/3
∴ (6+1)^th i.e. 7th term is the greatest term.

The ratio of the coefficients of x^p and x^q in the expansion of (1+x)^p+q is