Binomial Theorem & its Simple Applications Test - 4

 n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = ⁿ C_r . p^r . q^(n-r)
= ⁴⁴ C_r . (2y)^r . (-4x)^(44-r)

x⁴ can be achieved in the following ways: 
x⁴ . 1^(n-4) . (x² )⁰ . (x³ )⁰
Hence, coefficient will be  ⁿ C₄ .
x² . 1^(n-3) . (x² )¹ . (x3)⁰
Hence, coefficient will be 3ⁿ C₃ .
x¹ . 1^(n-2) . (x² )⁰ . (x³ )¹
Hence, coefficient will be 2ⁿ C₂ .
x⁰ . 1^(n-2) .(x² )² .(x³ )⁰
Hence, coefficient will be ⁿ C₂ .
Hence, the required coefficient will be  
ⁿ C₄ + 3ⁿ C₃ + 3ⁿ C₂
= ⁿ C₄ + 3(ⁿ C₃ + ⁿ C₂ ).
= ⁿ C₄  + 3(ⁿ⁺¹ C₃ ).
= ⁿ C₄  + ⁿ C₂ + ⁿ C₁ . ⁿ C₂

 Largest coefficient in the expansion of (1+x)²⁴  
= ²⁴ C_24/2  
= ²⁴ C₁₂

we can write (6 ⁿ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ= 1 + nx + n(n-1)x ²/2! + n(n-1)(n-2)x ³/3! +.............∞use this here,
(6)ⁿ= (1 + 5)ⁿ= 1 + 5n + n(n-1)5 ²/2! + n(n-1)(n-2)5 ³/3! +...........∞
= 1 + 5n + 5 ²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}
Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞
6 ⁿ= 1 + 5n + 25P
6 ⁿ- 5n = 1 + 25P -------(1)
but we know, according to Euclid algorithm ,
dividend = divisor ×quotient + remainder ---(2)
compare eqn (1) to (2)
we observed that 6 ⁿ-5 n always leaves the remainder 1 when divided by 25

 (1+a)^m+n
Coefficient of am = m+nCm = m+n
Coefficient of aⁿ = m+nCn
nCx = nC(n −x) Property
Hence, coefficients of a^m and aⁿ are equal.

(126)1^1/3
=  (125 + 1)^1/3
=  [125 (1 + (1/125))]^1/3
=  125^1/3 (1 + (1/125))^1/3     1/125 <1
= 5 [1 + (1/3)(1/125) + ..........]
= 5 [1 + (1/3)(0.008)]
= 5 [1 + 0.002666]
= 5.013

[(1+x)/(1 −x)]²
⇒(1+x)² (1 −x)^-2
⇒(1 + x² + 2x)[1 + 2x + 3x² +..... +(n −1)x^n-2 + nx^n-1 + (n+1)xⁿ +...]
coeff of xn will be given by
(I) When 1 will be multiplied by (n+1)xⁿ
(II) When x2 will be multiplied by (n −1)x^n-1
(III) When 2x will be multiplied by nx^n-1
∴coeff. of xⁿ = n + 1 + n −1 + 2n
= 4n