JEE Advanced Level Test: Binomial Theorem- 3 - JEE MCQ

Coefficient of x⁵ in (1 + x)¹⁰ 
Coefficient of x²⁵ in (1 + x)³⁰

(n - 2r)² = n + 2

Factorise and expand

Consider the given function:
c1/c0 + 2c2/c1 +  3c3/c2 + 4c4/c3 +......+ncn/cn−1
=n/1 + [2n(n−1)]/2! . 1/n + [3n(n−1)(n−2)]/3! . 1/[n(n−1)/2!] +......+n.1/n
=n+(n−1)+(n−2)+......1
 =1+2+3+......+n
 = n(n+1)/2

As the value of ‘n’ increases, it becomes difficult and tedious to calculate the value of ⁿC_x.

[2(a³ + 3.a.16²)]²

Applying the above conditions, we get
nC2 + nC3+ nCn−2 + nCn−3 = 440
2(nC2 + nC3)=440
nC2 ​ + nC3=220
n(n−1)/2!+ [n(n−2)(n−1)]/3!=220
n(n−1)/2 ​(1+(n−2)/3=220
n(n−1)/2 (n+1)/3=220
n(n²−1)/6 ​= 220
n³−n=1320
n³−n−1320=0
The only real root of the above equation is 11.
Hence n=11

Given, G = (5√5 + 11)^{2n + 1} , 0 < G < 1 and R = (5√5 + 11)^{2n + 1}

Expand using the formula 

Expand (x + ai)ⁿ and (x – ai)ⁿ then multiply.

Take (1 + x)¹⁰⁰⁰ as common, after simplification it becomes (1 + x)¹⁰⁰⁰ coefficient of x⁵⁰ is ¹⁰⁰²C₅₀

2 + 4 + 8 + …..1024

Coefficient of

Correct Answer :- d

Explanation : x = (2 + (3)^1/2)ⁿ

For n = 1

x = 2 + (3)^1/2

f = x - [x]

=> [2 + (3)^1/2] - [2 + (3)^1/2] 

=> (3)^1/2 - 1

f² = [(3)^1/2 - 1]²

=> 2(2 - (3)^1/2)

1 - f = 1 - ((3)^1/2 - 1)

1 - f = (2 - (3)^1/2)

f²/(1-f) = [2(2-(3)^1/2)]/[(2-(3)^1/2)]

For n = 2

f = (7 + 4(3)^1/2) - (7 + 4(3)^1/2)

f = 7 + 4(3)^1/2 - 13

f = 4(3)^1/2 - 6

f² = [4(3)^1/2 - 6]^1/2

=> 48 + 36 - 48(3)^1/2

=> 84 - 48(3)^1/2

1 - f = 1 - (48(3)^1/2 - 6)

1 - f = 7 - 4(3)^1/2

f²/(1-f) = [84 - 48(3)^1/2]/[7 - 4(3)^1/2]

= 12(Even)

2²⁰⁰⁶ = 4(8)⁶⁶⁸ = 4 (7 + 1)⁶⁸⁸ leaves remainder 4 
2006 = 7 x 268 + 4 leaves remainder
∴ 2²⁰⁰⁶ - 2006 leaves remainder

For a discrete probability function, the variance is given by

Where µ is the mean, substitute P(x)=ⁿC_x p^x q^(n-x) in the above equation and put µ = np to obtain
V = npq

for this term to be independent of x. we must have


So, the required coefficient is

¹⁰C₄(-1)⁴ = 210

⁹⁵C₄ + ⁹⁵C₃ + ⁹⁶C₃ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁸C₄ + ⁹⁹C₃ 
= ⁹⁶C₄ + ⁹⁶C₃ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁹C₃
= ⁹⁷C₄ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁹C₃ 
= ⁹⁸C₄ + ⁹⁸C₃ + ⁹⁹C₃ =  ⁹⁹C₄ + ⁹⁹C₃ = ¹⁰⁰C₄ 

(2^1/5 +3^1/10)⁵⁵ is
Total term = 55 + 1 = 56

Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;  
Number of irrational terms = 56 - 6 = 50

Given expansion is 
∴ General term


Since, we have to find coefficient of x^-10 ∴ -12 + 2r = -10  ⇒ r = 1
Now, then  coefficient  of x ^-10 is 12 C₁(a)¹¹ (b)¹ = 12a¹¹b

Here 2n is even integer, therefore,

term  will be  the middle term.
Now, (n + 1)^th term in 

 

Replacing x by -x, 


Subtracting and putting

Given expression

Given expansion 



Number of dissimilar  term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1