Binomial Theorem & its Simple Applications Test - 2

(x+a)ⁿ = ⁿ C₀ xⁿ + ⁿ C₁ x^(n-1) a¹ + ⁿ C₂ x^(n-2) a² + ..........+ ⁿ C_(n-1) xa^(n-1) + ⁿ C_n  aⁿ
Now, ⁿ C₀ = ⁿ C_n , ⁿ C₁ = ⁿ C_n-1 ,   ⁿ C₂ = ⁿ C_n-2 ,........
therefore, ⁿ C_r = ⁿ C_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 ⁿ C₇ (2)^n-7 (1/3)⁷
coefficient of x^-8 in [2+1/3x]n is ⁿ C₈ (2)^n-8 (1/3)⁸
⟹ⁿ C₇ ×(2^n-7 ) ×1/3⁷ = ⁿ C₈ ×(2^n-8 ) ×1/3⁸
⟹ⁿ C₈ /ⁿ C₇ = 6
⟹(n-7)/8 = 6
⟹n = 55

General term T_r+1 of (x+y)ⁿ is given by  
T_r+1 = ⁿ C_r x^n-r y^r
T₂ = ⁿ C₂ x^n-2 y = 240
T₃ = ⁿ C₃ x^n-3 y² = 720
T₄ = ⁿ C₄ 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

We can expand  (1+x)(1 −x)ⁿ  as  
= (1+x)(ⁿ C₀ ​− ⁿ C₁ ​.x + ⁿ C₂ ​.x²   + ........ +(−1)^{n  n} C_n ​xⁿ )
Coefficient of  xⁿ  is  
(−1)^{n  n} C_n   ​+ (−1)^{n −1  n} C_{n −1}
​=(−1)ⁿ (1 −n)

x ²(1 + x)⁹⁸+ x ³(1 + x)⁹⁷+ ... + x ⁵⁴(1 + x)⁴⁶
It is a G.P. with first term = x ²(1 + x)⁹⁸
and common ratio = x / (1 + x)
sum of these terms = x ²(1 + x)⁹⁸* [( (x / (1 + x))⁵³- 1 ) / (x / (1 + x) - 1))]
= x ²(1 + x)⁹⁸* ((1 + x) - x ⁵³(1 + x)⁻⁵²)

= ⁹⁹C ₆₈- ⁴⁶C ₁₅
⇒p = 68, q = 15
⇒p + q = 83

Given, binomial expression is (y ²+ c / y)5  
Now, Tr+1 = 5Cr ×(y ²)5-r ×(c / y)r  
= 5Cr ×y10-3r ×Cr  
Now, 10 –3r = 1  
⇒3r = 9  
⇒r = 3  
So, the coefficient of y = 5C3 ×c ³= 10c ³

(√a x² + (1 / 2x³ ))¹⁰
T_r+1 = ¹⁰C_r (√a x² )^10-r (1 / 2x³ )^r
Independent of x ⇒20 - 2r - 3r = 0
r = 4
Independent of x is ¹⁰C ₄(√a)⁶ (1 / 2)⁴ = 105
(210 / (2 ×8)) a³ = 105
⇒a = 2
a² = 4

For constant term - 12 + r + r = 0
⇒r = 6
∴Constant term = 
= ¹²C ₆×(2 ⁶/ 25) ×3 ×3^1/5
= (231 / 25) ×2 ⁸×3^1/5 ×3
= (693 / 25) ×2 ⁸×⁵√3
∴α= 693 / 25
25 α= 693

⁵ C₀ (c/y)⁰ (y² )^5-0 + ⁵ C₁ (c/y)¹ (y² )^5-1 +⁵ C₂ (c/y)² (y² )^5-2 +....... + ⁵ C₅ (c/y)⁵ (y² )^5-5
∑(r = 0 to 5) ⁵ C_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 = ⁵ C₃ .C³
= 10C³

(1 + x)ⁿ= ⁿC ₀+ ⁿC ₁x ¹+ ⁿC ₂x ²+ ... + ⁿC ₙx ⁿ
ⁿC ₄, ⁿC ₅&ⁿC ₆are in A.P.
ⁿC ₅- ⁿC ₄= ⁿC ₆- ⁿC ₅
⇒n! / (5!(n - 5)!) - n! / (4!(n - 4)!) = n! / (6!(n - 6)!) - n! / (5!(n - 5)!)
⇒30(n - 9)(n - 6) = 5(n - 4)(n - 11)
⇒30n ²- 450n + 1620 = 5n ²
⇒1 / (n - 5) [ (n - 4 - 5) / (5(n - 4)) ] = 1 / 5 [ (n - 5 - 6) / (6(n - 5)) ]
⇒(n - 9) / (5(n - 4)) = 1 / 5 [ (n - 11) / 6 ]
⇒n ²- 21n + 98 = 0
n ₘₐₓ= 14

For rational terms,
r/3 and r/5 must be integer
3 and 5 divide r ⇒15 divides r ⇒r = 0 and r = 15
¹⁵C ₀5 ⁰2 ³+ ¹⁵C ₁₅5 ⁵2 ⁰
= 8 + 3125
= 3133

Given that 2 - p, p, 2 - α, αare four consecutive terms in the binomial expansion of (1 + x)ⁿ, they must follow the property of binomial coefficients:
Next term / Previous term = (n - r) / (r + 1)
Using this property:
(p / (2 - p)) = ((n - (r - 1)) / r)
((2 - α) / p) = ((n - r) / (r + 1))
(α/ (2 - α)) = ((n - (r + 1)) / (r + 2))
After solving these equations, it turns out that the given expression:
p ²- α²+ 6 α+ 2p
simplifies to 8.
Thus, the correct option is: A. 8.

n-1C_r = (k ²- 8) nC_r+1

(^n-1 C_r ) / (ⁿ C_r+1 ) = k ²- 8
(r + 1) / n = k ²- 8
⇒k ²- 8 >0
(k - 2 √2)(k + 2 √2) >0
k ∈(-∞, -2 √2) ∪(2 √2, ∞) .... (I)
∴n ≥r + 1, (r + 1) / n ≤1
⇒k ²- 8 ≤1
k ²- 9 ≤0
-3 ≤k ≤3 .... (II)
From equation (I) and (II) we get
k ∈[-3, -2 √2) ∪(2 √2, 3]

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

Sum of coefficients in the expansion of (1 - 3x + 10x ²)ⁿ= A
then A = (1 - 3 + 10)ⁿ= 8 ⁿ(put x = 1)
and sum of coefficients in the expansion of
(1 + x ²)ⁿ= B
then B = (1 + 1)ⁿ= 2 ⁿ
A = B ³

T₂ = _n C¹ y¹ x^(n-1) = 135
T₃ = _n C² y² x^(n-2) = 30
T₄ = _n C³ y³ x^(n-3) = 10/3
⇒135/30 = (x/y) * n * 2 / n(n-1) = (2 / n-1) * (x/y) ... (i)
30 / (10/3) = n(n-1) / 2 / n(n-1)(n-2) * 3! * (x/y)
9 = (3 / n-2) * (x/y)
3(n - 2) = 135 / 60 (n - 1) ⇒n = 5
⇒x = 9y ... (i)
y * x⁴ = 27 ⇒x / 9 * x⁴ = 3³
⇒x⁵ = 3⁵ ⇒x = 3y = 1/3
⇒6 (5³ + 3² + 1/3) = 6 (125 + 9 + 1/3)
= 6(134) + 2 = 806

Given, (2x ³- (1 / 3x ²))⁵
General term,

∴15 - 5r = 5
∴r = 2
T ₃= 10 (8 / 9) x ⁵
So, coefficient is 80 / 9.

General term of (3/2 x² - 1/3 x)⁹
T_r+1 = ⁹ C_r (3/2 x² )^(9-r) (-1/3x)^r = ⁹ C_r (-1)^r 3^9-2r 2^r-9 x^18-3
Constant term in expansion of (1 + 2x - 3x³ )
(3/2 x² - 1/3 x)⁹
= T₇ - T₈ = ⁹ C₆ 3^(-3) 2^(-3) + 3⁹ C₇ 3^(-5) 2^(-2)
= 3 ×4 ×7 / 33 * 23 + 3 ×9 ×4 / 35 ×22
p = 54 / 108
108p = 54

{ (4²⁰²² ) / 15 } = { (2⁴⁰⁴⁴ ) / 15 }
= { (1 + 15)¹⁰¹¹ / 15 }
= 1 / 15

(x + 3)^(n-1) + (x + 3)^(n-2) (x + 2) + (x + 3)^(n-3) (x + 2)² + ... + (x + 2)^(n-1)
∑αr = 4^(n-1) + 4^(n-2) ×3 + 4^(n-3) ×3² + ... = 4^(n-1) [1 + 3/4 + (3/4)² + ... + (3/4)^(n-1) ]
= 4^(n-1) ×(1 - (3/4)ⁿ ) / (1 - 3/4)
= 4^(n-1) ×4 ×(1 - (3/4)ⁿ )
= 4^(n-1) ×4 ×(βⁿ - γⁿ )
β= 4, γ= 3
β² + γ² = 16 + 9 = 25

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

For coefficient of x^2/3
⇒6 - (16r/15) = 2/3
⇒90 - 16r = 10
⇒r = 5
For coefficient of x^-2/5
⇒6 - (16r/15) = -2/5
⇒90 - 16r = -6
⇒r = 6
Sum of coefficient of x^2/3 &x^-2/5
= ⁹C ₅. (1/2⁵ ) + ⁹C ₆. (1/2⁶ )
= (9! / 5!4!) x (1/2⁵ ) + (9! / 6!3!) x (1/2⁶ ) = 21/4

(¹⁰⁰C ₀- ¹⁰⁰C ₁+ ¹⁰⁰C ₂- ... - ¹⁰⁰C ₄₉) + ¹⁰⁰C ₅₀
+ (-¹⁰⁰C ₅₁+ ¹⁰⁰C ₅₂- ... + ¹⁰⁰C ₁₀₀) = 0
λ₁+ ¹⁰⁰C ₅₀+ λ₂= 0
λ₁= - (1/2) ¹⁰⁰C ₅₀(∵λ₁= λ₂)
= -⁹⁹C ₄₉

428 = 21 ×20 + 8
⇒428²⁰²⁴ ≡(20 ×21 + 8)²⁰²⁴ ≡8²⁰²⁴ (mod 21)
8² = 21 ×3 + 1
8²⁰²⁴ = (21 ×3 + 1)¹⁰¹²
⇒8²⁰²⁴ ≡(21 ×3 + 1)¹⁰¹² (mod 21)
≡1²⁰¹² (mod 21)
428²⁰²⁴ ≡1 (mod 21)

General term in expansion of (7^1/2 + 11^1/6 )⁸²⁴ is T_(r+1) = ⁸²⁴ C_r * 7^((824-r)/2) * 11^(r/6)
For integral term, r must be a multiple of 6.
Hence r = 0, 6, 12, ..., 822