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All questions of Binomial Theorem & its Simple Applications for JEE Exam
In the expansion of the binomial expansion (a + b)n, which of the following is incorrect ?
- a)The exponent of a decreases from n to 0.
- b)The number of terms is n+1.
- c)The exponent of b increases from 0 to n.
- d)The binomial coefficients in (a+b)n, which are equidistant from beginning and end are not equal.
Correct answer is option 'D'. Can you explain this answer?
In the expansion of the binomial expansion (a + b)n, which of the following is incorrect ?
a)
The exponent of a decreases from n to 0.
b)
The number of terms is n+1.
c)
The exponent of b increases from 0 to n.
d)
The binomial coefficients in (a+b)n, which are equidistant from beginning and end are not equal.
|
Krishna Iyer answered |
Correct Answer: d
Explanation:- The coefficient of terms (x+a)n equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficients.
nCr = nCn – r, r = 0,1,2,…,n.
The middle term in the expansion of (x + y)10 is the- a)6th term
- b)5th term
- c)average of 5th and 6th terms
- d)7th term
Correct answer is option 'A'. Can you explain this answer?
The middle term in the expansion of (x + y)10 is the
a)
6th term
b)
5th term
c)
average of 5th and 6th terms
d)
7th term
|
Vikas Kapoor answered |
Number of terms(n) = 10
Middle term = (n/2) + 1
Middle term = (n/2) + 1
= (10/2) + 1
= 5 + 1
= 6th term
= 5 + 1
= 6th term
The largest coefficient in the expansion of (1+x)24 is:- a)24C13
- b)24C12
- c)24C11
- d)24C24
Correct answer is option 'B'. Can you explain this answer?
The largest coefficient in the expansion of (1+x)24 is:
a)
24C13
b)
24C12
c)
24C11
d)
24C
24
|
Manish Aggarwal answered |
Largest coefficient in the expansion of (1+x)24
= 24C24/2
= 24C12
= 24C24/2
= 24C12
The middle term in the expansion of (1 – 2x + x2)n is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The middle term in the expansion of (1 – 2x + x2)n is
a)
b)
c)
d)
|
Om Desai answered |
Here 2n is even integer, therefore,
term will be the middle term.
Now, ( n + 1)th term in (1-x)2n = 2nCn(1)2n-n(-x)n = 2nCn(-x)n
The number of terms in the expansion of (x – y + 2z)7 are:- a)35
- b)38
- c)36
- d)37
Correct answer is option 'C'. Can you explain this answer?
The number of terms in the expansion of (x – y + 2z)7 are:
a)
35
b)
38
c)
36
d)
37
|
Tejas Verma answered |
Here the number of terms can be calculated by:
= ((n+ 1) * (n+2)) /2
where, n =7
= ((n+ 1) * (n+2)) /2
where, n =7
∴ Number of terms = 36
is
The number of terms in the expansion of (2x - 3y)8 is- a)6
- b)9
- c)8
- d)5
Correct answer is option 'B'. Can you explain this answer?
The number of terms in the expansion of (2x - 3y)8 is
a)
6
b)
9
c)
8
d)
5
|
Aryan Khanna answered |
Since this binomial is to the power 8, there will be nine terms in the expansion.
are equal, then n is equal to
The sum 
equals - a)nCr+1
- b)n+1Cr+1
- c)n+1Cr-1
- d)n+1Cr
Correct answer is option 'B'. Can you explain this answer?
The sum equals
equals a)
nCr+1
b)
n+1Cr+1
c)
n+1Cr-1
d)
n+1Cr
|
Ishan Choudhury answered |
C(n, r) + c(n -1, r) + C(n - 2, r) + . . . + C(r, r)
= r+1Cr+1 + r+1Cr + r+2Cr + . . . . + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again)
(∴ nCr + nCr-1 = n+1Cr)
= r+1Cr+1 + r+1Cr + r+2Cr + . . . . + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again)
(∴ nCr + nCr-1 = n+1Cr)
If 'a' be the sum of the odd terms & 'b' the sum of the even terms in the expansion of (1+x)n, then (1–x2)n equals- a)a2 – b2
- b)a2 + b2
- c)b2 – a2
- d)None
Correct answer is option 'A'. Can you explain this answer?
If 'a' be the sum of the odd terms & 'b' the sum of the even terms in the expansion of (1+x)n, then (1–x2)n equals
a)
a2 – b2
b)
a2 + b2
c)
b2 – a2
d)
None
|
Riya Banerjee answered |
Sum of odd terms in the expansion of (1+x)n will be 2(n-1) ....(i)
Now
(1−x2)n = nC0 − nC1 x2 + nC2 x4 + ....(−1)n nCn x2n .....(n)
(1+x2)n = nC0 + nC1 x2 + nC2 x4 +... nCn x2n .....(m)
Subtracting n from m, we get
(1+x2)n − (1−x2)n = 2[nC1 x2 + nC3 x6 + .....]
Now let x=1.
Hence we get
2n − 0 = 2[nC1 + nC3 + .....]
Or nC1 + nC3x2 + nC5 + ... = 2(n-1)
Hence a = b
Now (1−x2)n
=(1+x)n (1−x)n
→[(a+b)(a−b)]
=a2 − b2.
Now
(1−x2)n = nC0 − nC1 x2 + nC2 x4 + ....(−1)n nCn x2n .....(n)
(1+x2)n = nC0 + nC1 x2 + nC2 x4 +... nCn x2n .....(m)
Subtracting n from m, we get
(1+x2)n − (1−x2)n = 2[nC1 x2 + nC3 x6 + .....]
Now let x=1.
Hence we get
2n − 0 = 2[nC1 + nC3 + .....]
Or nC1 + nC3x2 + nC5 + ... = 2(n-1)
Hence a = b
Now (1−x2)n
=(1+x)n (1−x)n
→[(a+b)(a−b)]
=a2 − b2.
In the expansion of (a+b)n, nN the number of terms is:- a)n + 1
- b)n
- c)n – 1
- d)an
Correct answer is option 'A'. Can you explain this answer?
In the expansion of (a+b)n, nN the number of terms is:
a)
n + 1
b)
n
c)
n – 1
d)
an
|
Tanishq Saini answered |
Explanation:
Binomial Expansion:
In the expansion of (a+b)^n, where n is a positive integer, the number of terms in the expansion is (n+1).
Proof:
When we expand (a+b)^n using binomial theorem, the general term of the expansion is given by:
T(r+1) = nCr * a^(n-r) * b^r
where r is the index of the term and nCr represents the binomial coefficient.
Number of Terms:
To find the number of terms in the expansion, we need to find the values of r that satisfy 0 ≤ r ≤ n.
Counting the Terms:
- When r = 0, we get the first term: nC0 * a^n * b^0 = a^n
- When r = 1, we get the second term: nC1 * a^(n-1) * b^1
- ...
- When r = n, we get the (n+1)th term: nCn * a^0 * b^n = b^n
Total Number of Terms:
Since we have considered all values of r from 0 to n, there are (n+1) terms in the expansion of (a+b)^n.
Therefore, the correct answer is option A) n + 1.
Binomial Expansion:
In the expansion of (a+b)^n, where n is a positive integer, the number of terms in the expansion is (n+1).
Proof:
When we expand (a+b)^n using binomial theorem, the general term of the expansion is given by:
T(r+1) = nCr * a^(n-r) * b^r
where r is the index of the term and nCr represents the binomial coefficient.
Number of Terms:
To find the number of terms in the expansion, we need to find the values of r that satisfy 0 ≤ r ≤ n.
Counting the Terms:
- When r = 0, we get the first term: nC0 * a^n * b^0 = a^n
- When r = 1, we get the second term: nC1 * a^(n-1) * b^1
- ...
- When r = n, we get the (n+1)th term: nCn * a^0 * b^n = b^n
Total Number of Terms:
Since we have considered all values of r from 0 to n, there are (n+1) terms in the expansion of (a+b)^n.
Therefore, the correct answer is option A) n + 1.
The expansion of 
, in powers of x, is valid if- a)|x| > 2
- b)x < 2
- c)x > 2
- d)|x| < 2
Correct answer is option 'D'. Can you explain this answer?
The expansion of , in powers of x, is valid if
, in powers of x, is valid ifa)
|x| > 2
b)
x < 2
c)
x > 2
d)
|x| < 2
|
Knowledge Hub answered |
In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are- a)additive inverse of each other
- b)multiplicative inverse of each other
- c)equal
- d)nothing can be said
Correct answer is option 'C'. Can you explain this answer?
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are
a)
additive inverse of each other
b)
multiplicative inverse of each other
c)
equal
d)
nothing can be said
|
Hansa Sharma answered |
(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn an
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
The value of 1261/3 upto three decimals is- a)5.013
- b)5.014
- c)5.012
- d)5.011
Correct answer is option 'A'. Can you explain this answer?
The value of 1261/3 upto three decimals is
a)
5.013
b)
5.014
c)
5.012
d)
5.011
|
Lavanya Menon answered |
(126)11/3
= (125 + 1)1/3
= [125 (1 + (1/125))]1/3
= 1251/3(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
= (125 + 1)1/3
= [125 (1 + (1/125))]1/3
= 1251/3(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
- a)
- b)
- c)610
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
610
d)
|
|
Om Desai answered |
(1- x+2x2)10 = a0+a1x+a2x2 +..a19x19 + a20x20
Replacing x by-x, (1+x+ 2x2)10 = a0 - 1x+a2x2...a19x19 +a20x20)
Subtracting and putting x = 1, 210 - 410 = 2(a1 +a3 +...+a19)
⇒
Replacing x by-x, (1+x+ 2x2)10 = a0 - 1x+a2x2...a19x19 +a20x20)
Subtracting and putting x = 1, 210 - 410 = 2(a1 +a3 +...+a19)
⇒
the coefficient of xn in the expansion of 
- a)n
- b)2n
- c)4n
- d)zero
Correct answer is option 'C'. Can you explain this answer?
the coefficient of xn in the expansion of
a)
n
b)
2n
c)
4n
d)
zero
|
Suresh Reddy answered |
[(1+x)/(1−x)]2
⇒ (1+x)2(1−x)-2
⇒ (1 + x2 + 2x)[1 + 2x + 3x2 +..... +(n−1)xn-2 + nxn-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)xn-1
(III) When 2x will be multiplied by nxn-1
∴ coeff. of xn = n + 1 + n − 1 + 2n
= 4n
⇒ (1+x)2(1−x)-2
⇒ (1 + x2 + 2x)[1 + 2x + 3x2 +..... +(n−1)xn-2 + nxn-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)xn-1
(III) When 2x will be multiplied by nxn-1
∴ coeff. of xn = n + 1 + n − 1 + 2n
= 4n
The coefficient of y in the expansion of (y² + c/y)5 is - a)10c
- b)10c²
- c)10c³
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The coefficient of y in the expansion of (y² + c/y)5 is
a)
10c
b)
10c²
c)
10c³
d)
None of these
|
|
Varun Kapoor answered |
Given, binomial expression is (y² + c / y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
Let n be a positive integer. Then of the following, the greatest term is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Let n be a positive integer. Then of the following, the greatest term is
a)
b)
c)
d)
|
Neha Joshi answered |
i)(1+1/4n)4n
= (1+1/4)4
= (5/4)4 = 625/256
ii)(1+1/3n)3n
= (1+1/3)3
= (4/3)3 = 64/27
iii)(1+1/2n)2n
=(1+1/2)2 = (3/2)2
= 6/4 = 3/2
iv)(1+1/n)n
=(1+1)1 = 2
= (1+1/4)4
= (5/4)4 = 625/256
ii)(1+1/3n)3n
= (1+1/3)3
= (4/3)3 = 64/27
iii)(1+1/2n)2n
=(1+1/2)2 = (3/2)2
= 6/4 = 3/2
iv)(1+1/n)n
=(1+1)1 = 2
The number of dissimilar terms in the expansion of (a + b + c)2n+1 – (a + b – c)2n+1 is- a)(n + 1)2
- b)(n – 1)2
- c)4n2 – 1
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of dissimilar terms in the expansion of (a + b + c)2n+1 – (a + b – c)2n+1 is
a)
(n + 1)2
b)
(n – 1)2
c)
4n2 – 1
d)
none of these
|
|
Vikas Kapoor answered |
Given expansion
Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1
Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1
In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by- a)np
- b)npq
- c)np2q
- d)npq2
Correct answer is option 'B'. Can you explain this answer?
In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by
a)
np
b)
npq
c)
np2q
d)
npq2
|
Mehul Chavan answered |
For a discrete probability function, the variance is given by
Where µ is the mean, substitute P(x)=nCx px q(n-x) in the above equation and put µ = np to obtain
V = npq
Where µ is the mean, substitute P(x)=nCx px q(n-x) in the above equation and put µ = np to obtain
V = npq
What is the coefficient of x5 in the expansion of (1-x)-6 ?- a)252
- b)250
- c)-252
- d)251
Correct answer is option 'A'. Can you explain this answer?
What is the coefficient of x5 in the expansion of (1-x)-6 ?
a)
252
b)
250
c)
-252
d)
251
|
Preeti Iyer answered |
(1-x)-6
=> (1-x)(-6/1)
It is in the form of (1-x)(-p/q), p =6, q=1
=> (1-x)(-6/1)
It is in the form of (1-x)(-p/q), p =6, q=1
(1-x)(-p/q) = 1+p/1!(x/q)1 + p(p+q)/2!(x/q)2 + p(p+q)(p+2q)/3!(x/q)3 + p(p+q)(p+2q)(p+3q)/4!(x/q)4........
= 1+6/1!(x/1)1 + 6(7)/2!(x/1)2 + 6(7)(8)/3!(x/1)3 + 6(7)(8)(9)/4!(x/1)4 +.......................
So, coefficient of x5 is (6*7*8*9*10)/120
= 252
= 252
If in the expansion of (1+x)20, the coefficients of rth and (r+4)th terms are equal, then the value of r is equal to:- a)9
- b)7
- c)10
- d)8
Correct answer is option 'A'. Can you explain this answer?
If in the expansion of (1+x)20, the coefficients of rth and (r+4)th terms are equal, then the value of r is equal to:
a)
9
b)
7
c)
10
d)
8
|
|
Raghav Bansal answered |
Coefficients of the rth and (r+4)th terms in the given expansion are Cr−120 and 20Cr+3.
Here,Cr−120 = 20Cr+3
⇒ r−1+r+3 = 20
[∵ if nCx = nCy ⇒ x = y or x+y = n]
Here,Cr−120 = 20Cr+3
⇒ r−1+r+3 = 20
[∵ if nCx = nCy ⇒ x = y or x+y = n]
⇒ r = 2 or 2r = 18
⇒ r = 9
⇒ r = 9
What is the general term in the expansion of (2y-4x)44?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
What is the general term in the expansion of (2y-4x)44?
a)
b)
c)
d)
|
|
Om Desai answered |
n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = nCr . pr . q(n-r)
= 44Cr . (2y)r . (-4x)(44-r)
General term of (p+q)n is given by
T(r+1) = nCr . pr . q(n-r)
= 44Cr . (2y)r . (-4x)(44-r)
The coefficient of x4 in the expansion of (1 + x + x2 + x3)n is:
- a)nC4 + nC2
- b)nC4 + nC2+ nC4.nC2
- c)nC4 + nC2+ nC1.nC2
- d)nC4
Correct answer is option 'C'. Can you explain this answer?
The coefficient of x4 in the expansion of (1 + x + x2 + x3)n is:
a)
nC4 + nC2
b)
nC4 + nC2+ nC4.nC2
c)
nC4 + nC2+ nC1.nC2
d)
nC4
|
Gaurav Kumar answered |
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
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
Which of the following is divisible by 25:- a)6n - 5n + 1
- b)6n + 5n
- c)6n - 5n
- d)6n - 5n - 1
Correct answer is option 'D'. Can you explain this answer?
Which of the following is divisible by 25:
a)
6n - 5n + 1
b)
6n + 5n
c)
6n - 5n
d)
6n - 5n - 1
|
|
Riya Banerjee answered |
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
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
If the coefficients of x7 & x8 in the expansion of 
are equal, then the value of n is- a)15
- b)45
- c)55
- d)56
Correct answer is option 'C'. Can you explain this answer?
If the coefficients of x7 & x8 in the expansion of are equal, then the value of n is
are equal, then the value of n isa)
15
b)
45
c)
55
d)
56
|
|
Ananya Das answered |
Since Tr + 1 = nCr an – r xr in expansion of (a + x)n,
Therefore,
Therefore,
The general term in the expansion of (a - b)n is- a)Tr+1 = (-1)r nCr an-r br
- b)Tr+1 = nCr an-r br
- c)Tr+1 = nCr ar br
- d)Tr+1 = (-1)r nCr ar br
Correct answer is option 'A'. Can you explain this answer?
The general term in the expansion of (a - b)n is
a)
Tr+1 = (-1)r nCr an-r br
b)
Tr+1 = nCr an-r br
c)
Tr+1 = nCr ar br
d)
Tr+1 = (-1)r nCr ar br
|
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Poonam Reddy answered |
If a and b are real numbers and n is a positive integer, then:
(a - b)n = nC0 an + nC1 a(n – 1) b1 + nC2 a(n – 2) b2+ ...... + nCr a(n – r) br+ ... + nCnbn,
(a - b)n = nC0 an + nC1 a(n – 1) b1 + nC2 a(n – 2) b2+ ...... + nCr a(n – r) br+ ... + nCnbn,
The general term or (r + 1)th term in the expansion is given by:
Tr + 1 = (-1)Cr a(n–r) br
Tr + 1 = (-1)Cr a(n–r) br
If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is- a)15
- b)20
- c)5
- d)10
Correct answer is option 'C'. Can you explain this answer?
If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is
a)
15
b)
20
c)
5
d)
10
|
Shiksha Academy answered |
General term Tr+1 of (x+y)n is given by
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-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
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-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
n-1Cr = (k2 - 3). nCr + 1 [JEE 2004 (Scr.)]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
n-1Cr = (k2 - 3). nCr + 1 [JEE 2004 (Scr.)]
a)
b)
c)
d)
|
|
Rohit Jain answered |
(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
hence, k (√3, 2]
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
hence, k (√3, 2]
where n ∈ N and p ∈ N and 0 < f < 1 then the value, f2 – f + pf – p is
The coefficient of x3 in the binomial expansion of 
- a)792m5
- b)942m7
- c)330m4
- d)792m6
Correct answer is option 'C'. Can you explain this answer?
The coefficient of x3 in the binomial expansion of
a)
792m5
b)
942m7
c)
330m4
d)
792m6
|
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Geetika Shah answered |
T(r+1) = 11Cr x(11-2r) (m/r)r (-1)r
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4
The total number of 9 digit numbers which have all different digits is- a)9.9!
- b)10!
- c)10P9
- d)99
Correct answer is option 'A'. Can you explain this answer?
The total number of 9 digit numbers which have all different digits is
a)
9.9!
b)
10!
c)
10P9
d)
99
|
|
Geetika Shah answered |
Total number of digits =10
i.e. 0,1,2,3,4,5,6,7,8,9
require 9 diffrent number
0 cant be placed in first place
= first place can be filled in 9 ways
and the rest 9 blank with 9 digits in 9! ways
total ways = 9×9!
i.e. 0,1,2,3,4,5,6,7,8,9
require 9 diffrent number
0 cant be placed in first place
= first place can be filled in 9 ways
and the rest 9 blank with 9 digits in 9! ways
total ways = 9×9!
Fractional part of 
- a)2/31
- b)4/31
- c)8/31
- d)10/31
Correct answer is option 'C'. Can you explain this answer?
Fractional part of
a)
2/31
b)
4/31
c)
8/31
d)
10/31
|
|
Aryan Khanna answered |
= 8(1+31)15 = 8{15C0 + 15C131+..+15C15(31)15}
278 = 8 + an integer multiple of 31;
The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is- a)- 171
- b)342
- c)171/2
- d)684
Correct answer is option 'A'. Can you explain this answer?
The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is
a)
- 171
b)
342
c)
171/2
d)
684
|
Infinity Academy answered |
The coefficient of x17 is given by
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171
(a) Coefficient of t24 in the expansion of (1 + t2)12 (1 + t12) (1 + t24) is [JEE 2003 (Scr.), 3](b) Prove that :
[JEE 2003 (Mains),2]- a)12C6 + 2
- b)12C6+1
- c) 12C6
- d)None
Correct answer is option 'A'. Can you explain this answer?
(a) Coefficient of t24 in the expansion of (1 + t2)12 (1 + t12) (1 + t24) is [JEE 2003 (Scr.), 3]
(b) Prove that :
[JEE 2003 (Mains),2]a)
12C6 + 2
b)
12C6+1
c)
12C6
d)
None
|
|
Om Desai answered |
(1+x)12 (1+x12) (1+x24)
= [C0 + C1x2 + C2x4 + C3x6 + C4x8 +.......C12x24) (1 + x12 + x24 + x36)
= x24(12C0 + 12C6 + 12C12)
= 1 + 12C6 + 1
= 12C6 + 2
= [C0 + C1x2 + C2x4 + C3x6 + C4x8 +.......C12x24) (1 + x12 + x24 + x36)
= x24(12C0 + 12C6 + 12C12)
= 1 + 12C6 + 1
= 12C6 + 2
is, where 
= nCr [JEE 2005 (Scr.)]
In the expansion of (1+x)60, the sum of coefficients of odd powers of x is- a)261
- b)260
- c)259
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
In the expansion of (1+x)60, the sum of coefficients of odd powers of x is
a)
261
b)
260
c)
2
59
d)
none of these.
|
|
Mohit Mittal answered |
C is correct as
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59
The sum of coefficients of (1 + x - 3x2)2134 is- a)-1
- b)1
- c)0
- d)22134
Correct answer is option 'B'. Can you explain this answer?
The sum of coefficients of (1 + x - 3x2)2134 is
a)
-1
b)
1
c)
0
d)
22134
|
|
Anjana Sharma answered |
Sum of coefficients in (1+x−3x^2)^2143
⇒(1+1−3(1)^2)^2143
⇒(−1)^2143 = −1
Hence (A) is the correct answer.
Chapter doubts & questions for Binomial Theorem & its Simple Applications - Chapter-wise Tests for JEE Main & Advanced 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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