Test: General Terms In Binomial Expansion - JEE MCQ

 n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = ⁿC_r . p^r . q^(n-r)
= ⁴⁴C_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  ⁿC₄ .
x² . 1^(n-3) . (x²)¹ . (x3)⁰
Hence, coefficient will be 3ⁿC₃.
x¹ . 1^(n-2) . (x²)⁰ . (x³)¹
Hence, coefficient will be 2ⁿC₂.
x⁰ . 1^(n-2) .(x²)² .(x³)⁰
Hence, coefficient will be ⁿC₂ .
Hence, the required coefficient will be 
ⁿC₄ + 3ⁿC₃ + 3ⁿC₂
= ⁿC₄ + 3(ⁿC₃ + ⁿC₂).
= ⁿC₄  + 3(ⁿ⁺¹C₃).
= ⁿC₄  + ⁿC₂ + ⁿC₁ . ⁿC₂

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

we can write (6ⁿ ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! +.............∞ use this here,
(6)ⁿ = (1 + 5)ⁿ = 1 + 5n + n(n-1)5²/2! + n(n-1)(n-2)5³/3! +...........∞
= 1 + 5n + 5²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}
Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞
6ⁿ = 1 + 5n + 25P
6ⁿ - 5n = 1 + 25P -------(1)
but we know, according to Euclid algorithm ,
dividend = divisor × quotient + remainder ---(2)
compare eqn (1) to (2)
we observed that 6ⁿ -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