JEE Mathematics Binomial Theorem MCQs Set 09

Multiple Choice Questions (MCQs) for JEE Mathematics: Binomial Theorem

Access targeted multiple-choice questions for Binomial Theorem designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Binomial Theorem MCQs for JEE Mathematics

Access the complete set of multiple-choice questions for Binomial Theorem below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The term independent of x in the expansion of \( \left(\sqrt{\frac x3}+\frac3{2x^2}\right)^{10} \) is:
(a) \( \frac32 \)
(b) \( \frac54 \)
(c) \( \frac52 \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac54 \)

Explanation:
1. \( T_{r+1}={}^{10}C_r\left(\frac x3\right)^{\frac{10-r}2}\left(\frac3{2x^2}\right)^r \)
2. Power of x: \( \frac{10-r}2-2r=0\Rightarrow r=2 \)
3. \( T_3=45\cdot\left(\frac13\right)^4\cdot\frac94=\frac{45\cdot9}{81\cdot4}=\frac54 \)

Teacher's Note:
1. The root sign covers the whole fraction x over 3.
2. Set the exponent of x to zero to find r.

Question 2: Let the coefficients of \( x^n \) in \( (1+x)^{2n} \) and \( (1+x)^{2n-1} \) be P and Q respectively. Then \( \left(\frac{P+Q}Q\right)^5= \)
(a) 9
(b) 27
(c) 81
(d) none of these
Show Answer & Explanation

Answer: (d) none of these

Explanation:
1. \( P={}^{2n}C_n \), \( Q={}^{2n-1}C_n \)
2. \( \frac PQ=\frac{(2n)!}{n!\,n!}\cdot\frac{n!\,(n-1)!}{(2n-1)!}=\frac{2n}{n}=2 \)
3. \( \frac{P+Q}Q=3 \), so the value is \( 3^5=243 \), which is not among the listed values

Teacher's Note:
1. Express both coefficients with factorials and take the ratio.
2. Raise the simple ratio to the required power.

Question 3: The coefficient of \( x^{52} \) in the expansion \( \sum_{m=0}^{100}{}^{100}C_m(x-3)^{100-m}2^m \) is:
(a) \( {}^{100}C_{47} \)
(b) \( {}^{100}C_{48} \)
(c) \( -{}^{100}C_{52} \)
(d) \( -{}^{100}C_{100} \)
Show Answer & Explanation

Answer: (b) \( {}^{100}C_{48} \)

Explanation:
1. The sum is the binomial expansion of \( (x-3+2)^{100}=(x-1)^{100} \)
2. Coefficient of \( x^{52} \) \( ={}^{100}C_{52}(-1)^{48}={}^{100}C_{52}={}^{100}C_{48} \)

Teacher's Note:
1. Recognise the sum as a binomial expansion in a combined form.
2. Use nCr = nC(n-r).

Question 4: The coefficient of \( x^5 \) in the expansion of \( (1+x)^{21}+(1+x)^{22}+\dots+(1+x)^{30} \) is:
(a) \( {}^{51}C_5 \)
(b) \( {}^9C_5 \)
(c) \( {}^{31}C_6-{}^{21}C_6 \)
(d) \( {}^{30}C_5+{}^{20}C_5 \)
Show Answer & Explanation

Answer: (c) \( {}^{31}C_6-{}^{21}C_6 \)

Explanation:
1. The coefficient is \( \sum_{k=21}^{30}{}^kC_5 \)
2. By the hockey stick identity \( \sum_{k=5}^{30}{}^kC_5={}^{31}C_6 \) and \( \sum_{k=5}^{20}{}^kC_5={}^{21}C_6 \)
3. The required sum is \( {}^{31}C_6-{}^{21}C_6 \)

Teacher's Note:
1. Use the hockey stick identity to add binomial coefficients down a column.
2. Subtract the unwanted lower part.

Question 5: When \( (1+x)(1+x+x^2)(1+x+x^2+x^3)\cdots(1+x+x^2+\dots+x^{100}) \) is written in ascending powers of x, the highest exponent of x is:
(a) 5000
(b) 5030
(c) 5050
(d) 5040
Show Answer & Explanation

Answer: (c) 5050

Explanation:
1. The highest powers in the factors are \( 1,2,3,\dots,100 \)
2. Sum \( =\frac{100\cdot101}2=5050 \)

Teacher's Note:
1. The top exponent of a product is the sum of the top exponents.
2. Use the sum of the first 100 natural numbers.

Question 6: The numerically greatest term in the expansion of \( (2+3x)^9 \) when \( x=\frac32 \) is:
(a) \( {}^9C_6\cdot2^9\cdot\left(\frac32\right)^{12} \)
(b) \( {}^9C_3\cdot2^9\cdot\left(\frac32\right)^6 \)
(c) \( {}^9C_5\cdot2^9\cdot\left(\frac32\right)^{10} \)
(d) \( {}^9C_4\cdot2^9\cdot\left(\frac32\right)^8 \)
Show Answer & Explanation

Answer: (a) \( {}^9C_6\cdot2^9\cdot\left(\frac32\right)^{12} \)

Explanation:
1. \( (2+3x)^9=2^9\left(1+\frac{3x}2\right)^9 \) with \( \frac{3x}2=\frac94 \)
2. \( m=\left\lfloor\frac{n+1}{1+
3. a/b
4. }\right\rfloor=\left\lfloor\frac{10}{1+4/9}\right\rfloor=\left\lfloor6.92\right\rfloor=6 \), so the greatest term is \( T_7 \)
5. \( T_7={}^9C_6\,2^9\left(\frac94\right)^6={}^9C_6\,2^9\left(\frac32\right)^{12} \)

Teacher's Note:
1. Use the integral part of (n + 1)/(1 +
2. a/b
3. ) to locate the greatest term.
4. If the value is not an integer, the next term is the greatest.

Question 7: The remainder when \( 2^{2003} \) is divided by 17 is:
(a) 1
(b) 2
(c) 8
(d) none of these
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. \( 2^4=16\equiv-1\pmod{17} \), so \( 2^8\equiv1 \)
2. \( 2003=8\cdot250+3 \)
3. \( 2^{2003}\equiv2^3=8 \)

Teacher's Note:
1. Find the order of the base modulo 17.
2. Reduce the exponent by that order.

Question 8: The value of \( \binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50} \) is:
(a) \( \binom{100}{50} \)
(b) \( \binom{100}{51} \)
(c) \( \binom{50}{25} \)
(d) \( \binom{50}{25}^2 \)
Show Answer & Explanation

Answer: (b) \( \binom{100}{51} \)

Explanation:
1. \( \binom{50}{r}=\binom{50}{50-r} \), so the sum is \( \sum\binom{50}{50-r}\binom{50}{r+1} \)
2. By Vandermonde the coefficient sum is \( \binom{100}{51} \)

Teacher's Note:
1. Rewrite one factor with nCr = nC(n-r) to use Vandermonde's identity.
2. The lower indices then add to 51.

Question 9: The value of the expression \( \left(\sum_{r=0}^{10}{}^{10}C_r\right)\left(\sum_{K=0}^{10}(-1)^K\frac{{}^{10}C_K}{2^K}\right) \) is:
(a) \( 2^{10} \)
(b) \( 2^{20} \)
(c) 1
(d) \( 2^5 \)
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. First sum \( =2^{10} \)
2. Second sum \( =\left(1-\frac12\right)^{10}=2^{-10} \)
3. Product \( =1 \)

Teacher's Note:
1. Both sums are binomial expansions evaluated at special values.
2. The powers of 2 cancel.

Question 10: In the expansion of \( (1+x)^n\left(1+\frac1x\right)^n \), the term independent of x is:
(a) \( C_0^2+2C_1^2+\dots+(n+1)C_n^2 \)
(b) \( (C_0+C_1+\dots+C_n)^2 \)
(c) \( C_0^2+C_1^2+\dots+C_n^2 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( C_0^2+C_1^2+\dots+C_n^2 \)

Explanation:
1. The constant term pairs \( C_rx^r \) from the first factor with \( C_rx^{-r} \) from the second
2. So it is \( \sum C_r\cdot C_r \)

Teacher's Note:
1. A constant term needs equal and opposite powers of x.
2. Each pair contributes a square.

Question 11: If \( a_n=\sum_{r=0}^n\frac1{{}^nC_r} \), then \( \sum_{r=0}^n\frac{n-2r}{{}^nC_r} \) is:
(a) \( \frac n2a_n \)
(b) \( \frac14a_n \)
(c) \( na_n \)
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. Replace r by \( n-r \): the sum becomes \( \sum\frac{2r-n}{{}^nC_{n-r}}=-\sum\frac{n-2r}{{}^nC_r} \)
2. So the sum equals its own negative
3. Hence it is 0

Teacher's Note:
1. Reverse the order of summation.
2. A sum equal to its own negative is zero.

Question 12: The coefficient of \( a^5b^4c^7 \) in the expansion of \( (bc+ca+ab)^8 \) is:
(a) 180
(b) 240
(c) 280
(d) 32
Show Answer & Explanation

Answer: (c) 280

Explanation:
1. Take \( (bc)^p(ca)^q(ab)^r \) with \( p+q+r=8 \)
2. Powers: a gives \( q+r=5 \), b gives \( p+r=4 \), c gives \( p+q=7 \)
3. So \( p=3,q=4,r=1 \)
4. Coefficient \( =\frac{8!}{3!\,4!\,1!}=280 \)

Teacher's Note:
1. Write the exponents of each letter as equations.
2. Then apply the multinomial coefficient.

Question 13: The coefficient of \( x^4 \) in the expression \( (1+2x+3x^2+4x^3+\dots\infty)^{1/2} \), where \( |x|<1 \), is:
(a) 2
(b) 3
(c) 1
(d) 5
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. \( 1+2x+3x^2+\dots=(1-x)^{-2} \)
2. So the expression is \( \left[(1-x)^{-2}\right]^{1/2}=\frac1{1-x} \)
3. The coefficient of \( x^4 \) is 1

Teacher's Note:
1. Recognise the series as (1 - x) to the minus 2.
2. Taking the square root simplifies it.

Question 14: The sum of the infinite series \( \frac1{2!}+\frac{1+2}{3!}+\frac{1+2+3}{4!}+\dots \) to \( \infty \) is:
(a) \( \frac e3 \)
(b) e
(c) \( \frac e2 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac e2 \)

Explanation:
1. The nth term is \( \frac{n(n+1)/2}{(n+1)!}=\frac1{2(n-1)!} \)
2. Sum \( =\frac12\sum_{n\ge1}\frac1{(n-1)!}=\frac e2 \)

Teacher's Note:
1. Write the numerator as a sum of natural numbers.
2. Cancel to get a factorial series.

Question 15: In the expansion of \( \left(\sqrt[3]4+\frac1{\sqrt[4]6}\right)^{20} \):
(a) the number of irrational terms is 19
(b) the middle term is irrational
(c) the number of rational terms is 2
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. \( T_{r+1}={}^{20}C_r\,2^{\frac{2(20-r)}3}\,6^{-\frac r4} \)
2. Rational needs \( 3\mid(20-r) \) and \( 4\mid r \), so \( r=8 \) or \( 20 \): 2 rational terms
3. There are 21 terms, so 19 are irrational
4. The middle term is \( T_{11} \) with \( r=10 \), not a multiple of 4, so it is irrational
5. All three statements are true

Teacher's Note:
1. Write each radical with fractional exponents.
2. Test the divisibility conditions on r.

Question 16: If \( (1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\dots+a_{20}x^{20} \), then:
(a) \( a_1=20 \)
(b) \( a_2=210 \)
(c) \( a_4=8085 \)
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. \( a_1=10\cdot2=20 \)
2. \( a_2=10\cdot3+{}^{10}C_2\cdot2^2=30+180=210 \)
3. Counting the ways to get \( x^4 \) gives \( a_4=8085 \)
4. All three are correct

Teacher's Note:
1. Count the ways each power of x can arise from the factors.
2. Use the multinomial coefficients.

Question 17: In the expansion of \( \left(x^3-\frac1{x^2}\right)^n \), \( n\in\mathbb N \), if the sum of the coefficients of \( x^5 \) and \( x^{10} \) is 0, then n is:
(a) 25
(b) 20
(c) 15
(d) none of these
Show Answer & Explanation

Answer: (c) 15

Explanation:
1. \( T_{r+1}={}^nC_r(-1)^rx^{3n-5r} \)
2. For \( x^5 \): \( r=\frac{3n-5}5 \); for \( x^{10} \): \( r-1 \)
3. The coefficients are \( (-1)^r{}^nC_r \) and \( (-1)^{r-1}{}^nC_{r-1} \), which cancel when \( {}^nC_r={}^nC_{r-1} \), that is \( r+(r-1)=n \)
4. \( \frac{2(3n-5)}5-1=n\Rightarrow n=15 \)

Teacher's Note:
1. Find the term numbers for both powers.
2. Equal binomial coefficients occur when the lower indices add up to n.

Question 18: The term in the expansion of \( (2x-5)^6 \) which has the greatest binomial coefficient is:
(a) \( T_3 \)
(b) \( T_4 \)
(c) \( T_5 \)
(d) \( T_6 \)
Show Answer & Explanation

Answer: (b) \( T_4 \)

Explanation:
1. \( n=6 \) is even, so the greatest binomial coefficient is the middle one, \( {}^6C_3 \)
2. It belongs to \( T_4 \)

Teacher's Note:
1. The greatest binomial coefficient is at the middle.
2. Its term number is n/2 + 1 for even n.

Question 19: The last three digits of the number \( 27^{27} \) are:
(a) 805
(b) 301
(c) 503
(d) 803
Show Answer & Explanation

Answer: (d) 803

Explanation:
1. \( 27^3=19683\equiv683 \pmod{1000} \)
2. \( 27^9\equiv683^3\equiv987\equiv-13 \pmod{1000} \)
3. \( 27^{27}=(27^9)^3\equiv(-13)^3=-2197\equiv803 \pmod{1000} \)

Teacher's Note:
1. Reduce powers repeatedly modulo 1000.
2. Using a small negative residue such as -13 keeps the arithmetic short.

Question 20: \( 7^9+9^7 \) is divisible by:
(a) 7
(b) 24
(c) 64
(d) 72
Show Answer & Explanation

Answer: (c) 64

Explanation:
1. Write \( 7^9=(8-1)^9 \) and \( 9^7=(8+1)^7 \)
2. Modulo 64 only the first two terms matter: \( (8-1)^9\equiv-1+9\cdot8=71 \) and \( (8+1)^7\equiv1+7\cdot8=57 \)
3. The sum \( \equiv128\equiv0\pmod{64} \)
4. Direct division confirms \( 45136576=64\times705259 \), and it leaves non-zero remainders when divided by 7, 24 and 72

Teacher's Note:
1. Write the bases as 8 minus 1 and 8 plus 1 to see the factor 64.
2. Check the other options by their remainders.

Question 21: The coefficient of \( x^{n-1} \) in the expansion of \( (x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2+\dots+(x+2)^n \) is:
(a) \( {}^{n+1}C_2(3) \)
(b) \( {}^{n-1}C_2(5) \)
(c) \( {}^{n+1}C_2(5) \)
(d) \( {}^nC_2(5) \)
Show Answer & Explanation

Answer: (c) \( {}^{n+1}C_2(5) \)

Explanation:
1. The sum is a G.P. in \( \frac{x+2}{x+3} \): \( \frac{(x+3)^{n+1}-(x+2)^{n+1}}{(x+3)-(x+2)}=(x+3)^{n+1}-(x+2)^{n+1} \)
2. Coefficient of \( x^{n-1} \): \( {}^{n+1}C_2(3^2-2^2)={}^{n+1}C_2(5) \)

Teacher's Note:
1. Treat the sum as a geometric progression.
2. The denominator x + 3 - (x + 2) equals 1.

Free study material for Binomial Theorem

Practice MCQs for JEE Mathematics Binomial Theorem

Chapter MCQs with Answers for JEE Mathematics

Test your conceptual understanding of Binomial Theorem with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Mathematics, these problem sets build accuracy and prepare students for objective exams.

Expert Practice Material for JEE Mathematics

Built using the official NCERT book for JEE, these Mathematics objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Complete Your Chapter Revision

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Mathematics Binomial Theorem MCQs Set 09?

You can get most exhaustive JEE Mathematics Binomial Theorem MCQs Set 09 for free on StudiesToday.com. These MCQs for JEE Mathematics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics JEE material?

Yes, our JEE Mathematics Binomial Theorem MCQs Set 09 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in JEE exams?

By solving our JEE Mathematics Binomial Theorem MCQs Set 09, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for JEE Mathematics Binomial Theorem MCQs Set 09?

Yes, Mathematics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Mathematics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Mathematics Binomial Theorem MCQs Set 09 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.