Test: Binomial Theorem- 1 - 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

General term T_r+1 of (x+y)ⁿ is given by 
T_r+1 = ⁿC_r x^n-r y^r
T₂ = ⁿC₂ x^n-2 y = 240
T₃ = ⁿC₃ x^n-3 y² = 720
T₄ = ⁿC₄ x^n-4 y³ = 1080
T₃/T₂ = [(n-1)/2] * [y/x] = 3......(1)
T₄/T₂ = {[(n-1)(n-2)]/(3*2)} * x²/y² = 9/2
T₄/T₃ = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

Now, dividing (1) by (2) we get,





Now, Dividing (2) by (3) we get,

We can expand (1+x)(1−x)ⁿ as 
= (1+x)(ⁿC₀​− ⁿC₁​.x + ⁿC₂​.x² + ........ +(−1)^n nC_n​xⁿ)

Coefficient of xⁿ is 
(−1)^n nC_n ​+ (−1)^n−1 nC_n−1
​=(−1)ⁿ(1−n)

T_r+1 = ⁷⁴Cr₃ 3^73-r (2x)^r
T_r+2 = ⁷⁴C_r+1 3^73-r (2x)^r+1
 as co−efficients are equal 
⁷⁴Cr₃ 3^73-r 2^r = ⁷⁴C_r+1 3^73-r 2^r+1
 ((74!/74−r)!r!)×=(74!/(73−r)!(r+1)!) × 2
3(73−r)!(r+1)! = 2(74−r)!r!
= 3(73−r)!(r+1)! = 2(74−r)!(73−r)!r!
= 3(r+1) = 2(74−r)
= 3r+3 = 148−2r
∴ 5r = 148−3
r = (145/5) 
r = 29
Hence terms are (29+1)th and(29+2)nd or 30th and 31th term

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

⁵C₀ (c/y)⁰ (y²)^5-0 + ⁵C₁ (c/y)¹ (y²)^5-1 + ⁵C₂ (c/y)² (y²)^5-2 +....... + ⁵C₅ (c/y)⁵ (y²)^5-5
∑(r = 0 to 5) ⁵C_r (c/y)^r (y²)^5-r
We need coefficient of y ⇒2(5−r)−r=1
⇒ 10 − 3r = 1
⇒ r = 3
So, cofficient of y = ⁵C₃.C³
= 10C³

Co-efficient of x⁵ in (1+x²)⁵ (1+x)⁴
= {⁵C₀ (1)⁰ + ⁵C₁ (1)x² + ⁵C₂ (1)(x2)2 + ⁵C₃ (1)(x²)³ + ⁵C₄ (1)(x²)⁴ + ⁵C₅ (1)(x²)⁵} * {4C0 (1) + 4C1 (x) + 4C2 (x)2 + 4C3 (x)3 + 4C4 (x)4}
= {⁵C₂(x)⁴ * ⁴C₁(x) + ⁵C₁(x)⁴ + ⁵C₁(x)² * ⁴C₃(x)³} 
∴ coefficient of x5
=> (5×4)/(2×1) × 4)x⁵ + (5×4)x⁵
⇒(10×4+20)x⁵
⇒ 60x⁵

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCnxⁿ
Three consecutive terms coefficients
ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110
⇒  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!
ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!
ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!
6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!
⇒ 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!
⇒ 2(n-a) = 11(a + 1)
⇒ 2n - 2a = 11a + 11
⇒ 2n = 13a + 11
⇒ 13a = 2n - 11
33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!
⇒ 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!
⇒ 3 (n - a - 1) = 10(a + 2)
⇒ 3n - 3a - 3 = 10a + 20
⇒  3n = 13a + 23
⇒ 13a = 3n - 23
2n - 11 = 3n - 23
⇒ n = 12

5th term in the expansion of (x²/2 − 2/x²)¹² is
Tr+ 1= nCr x^r* y^(n−r)
T5 = 12C4[(x²/2)⁴ (-2/x²)⁸]
= (12! * x⁸ * 2⁸)/ (4! * 8! *2⁴ * x¹⁶)
= 7920x⁻⁴