CAT Question Bank  Binomial Theorem

Q9) Find the coefficient of the term independent of y in the expansion of [(y + 1)/( y^2/3 – y^1/3 + 1)  (y – 1) / (y – y^1/2)]^10

Q10) A man has 2n + 1 friends. The number of ways in which he can invite atleast n + 1 friends for a dinner is 4096. Find the number of friends of the man.

Q11) What is the highest power of 12 in (5^36)  1

Q12) What is d coefficient of x^204 in (x  1)(x^2  2)(x^3  3) ... (x^20  20)

Q13) The sum of the rational terms in (√2 + 3√3 + 6√6)^10 is
a. 12632
b. 12600
c. 126
d. None of these

Q14) Find coefficient of x^8 in (1 − 2x + 3x^2 − 4x^3 + 5x^4 − 6x^5 + 7x^6)^6

Q15) Number of terms in the expansion of (1 + x + x^2)^100 is

Q16) The number of irrational terms in the binomial expansion of (3^(1/5) + 7^(1/3))^100 is

Q17) In the expansion of (3^(x/4) + 3^(5x/4))^n, the sum of the binomial coefficients is 64 and the term with greatest binomial coefficient exceeds the third by (n1), then find x.

Q18) Find the coefficient of x^4 in (1+ 2x + 3x^2 + ... )^1/2

Q19)
a) Find the coefficient of x^11 in (1 + x + x^2 + x^3 + ... + x^20)^15
b) Find the coefficient of x^11 in (x + x^2 + x^3 + ... + x^20)^15
c) Find the coefficient of x^11 in (x^3 + x^4 + ... + x^20)^15

Q20) What is the highest power of 4 that divides 5^12  1 ?

Q21) A positive real number is increased by 10% and then decreased by 10%. This process is repeated 99 more times. What can be said about the new number obtained?
a. It is less than 30% of the original number.
b. It is between 30% and 45% of the original number.
c. It is between 45% and 55% of the original number.
d. It is greater than 55% of the original number

@rowdyrathore c is the ans


@rowdyrathore put x =1
(1+13)^2163 = (1)^2163 = 1

@rowdyrathore x=10

@badalravi
put logx = y and solve.
we will get 2y^2 + 3y = 5
solve the quadratic and the roots are y = 1 and 5/2
logx = 1 or logx = 5/2
x = 10 or 10^(5/2)
2 values possible.

@SoumyaRanjanPatra
nC4 * a^(n  4) * (b)^4 = nC5 * a^(n  5) * (b)^5
n! / 4! (n  4)! * a^(n  4) * b4  n!/5!(n  5)! * a^(n  5) * b^5 = 0
n!/4!(n  5)! * a^(n  5) * b4 [a/(n  4)  b/5] = 0
a/(n  4) = b/5
a/b = (n  4)/5
How you got 1 ?

@rowdyrathore 108