Binomial Theorem JEE Main PYQs (2020–2026)

Solved Binomial Theorem PYQs

3 recent questions with answers. The full set of 190 is free to practise.

1. JEE Main 2026EasyGeneral Term and Middle Term(s) in Binomial Expansion
   Q1

   The coefficient of $x^2$ in the expansion of ${\left(2x^2+\frac{1}{x}\right)}^{10},x\ne 0$ , is :

   • A3240
   • B3360
   • C3480
   • D3600

   Show answer & solution

   Correct answer: (B)

   Using the binomial theorem,
   
   $${\left(2x^2+\frac{1}{x}\right)}^{10}=\sum _{r=0}^{10}\left(\genfrac{}{}{0ex}{}{10}{r}\right)(2x^2{)}^{10-r}{\left(\frac{1}{x}\right)}^r$$
   
   Now simplify the general term:
   
   $$T_{r+1}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)2^{10-r}{x}^{2(10-r)}{x}^{-r}$$
   
   $$T_{r+1}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)2^{10-r}{x}^{20-2r-r}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)2^{10-r}{x}^{20-3r}$$
   
   We need the coefficient of $x^2$ , so set the power of $x$ equal to $2$ :
   
   $$20-3r=2$$
   
   $$3r=18$$
   
   $$r=6$$
   
   Now substitute $r=6$ :
   
   $$\text{Required coefficient}=\left(\genfrac{}{}{0ex}{}{10}{6}\right)2^{10-6}$$
   
   $$=\left(\genfrac{}{}{0ex}{}{10}{6}\right)2^4$$
   
   $$=210\times 16$$
   
   $$=3360$$
   
   So, the coefficient of $x^2$ is
   
   $$\boxed{3360}$$
   
   Hence, the correct option is B.
2. JEE Main 2026EasyTerm Independent of x & Rational/Irrational Terms
   Q2

   In the expansion of ${\left(9x-\frac{1}{3\sqrt{x}}\right)}^{18},x>0$, if the term independent of $x$ is (221)k, then k is equal to:

   • A84
   • B78
   • C168
   • D198

   Show answer & solution

   Correct answer: (A)

   For the expansion of
   
   $${\left(9x-\frac{1}{3\sqrt{x}}\right)}^{18},$$
   
   we use the general term in the binomial expansion:
   
   $$T_{r+1}=\left(\genfrac{}{}{0ex}{}{18}{r}\right)(9x{)}^{18-r}{\left(-\frac{1}{3\sqrt{x}}\right)}^r$$
   
   Now simplify the power of $x$ in this term.
   
   Since
   
   $$(9x{)}^{18-r}=9^{18-r}{x}^{18-r}$$
   
   and
   
   $${\left(\frac{1}{3\sqrt{x}}\right)}^r=\frac{1}{3^r{x}^{r/2}},$$
   
   so the power of $x$ in the general term is
   
   $$x^{18-r-\frac{r}{2}}=x^{18-\frac{3r}{2}}.$$
   
   For the term independent of $x$ , the power of $x$ must be zero:
   
   $$18-\frac{3r}{2}=0$$
   
   $$\frac{3r}{2}=18$$
   
   $$r=12$$
   
   So the required term is the $(12+1)$ th term, i.e. $T_{13}$ .
   
   Now compute it:
   
   $$T_{13}=\left(\genfrac{}{}{0ex}{}{18}{12}\right)(9x{)}^6{\left(-\frac{1}{3\sqrt{x}}\right)}^{12}$$
   
   Since the power of $x$ becomes zero, only the numerical coefficient remains:
   
   $$T_{13}=\left(\genfrac{}{}{0ex}{}{18}{12}\right)\cdot 9^6\cdot \frac{1}{3^{12}}$$
   
   Also,
   
   $$9^6=(3^2{)}^6=3^{12}$$
   
   So,
   
   $$T_{13}=\left(\genfrac{}{}{0ex}{}{18}{12}\right)$$
   
   and
   
   $$\left(\genfrac{}{}{0ex}{}{18}{12}\right)=\left(\genfrac{}{}{0ex}{}{18}{6}\right)=\frac{18!}{6!12!}=18564$$
   
   Given that the term independent of $x$ is $(221)k$ , we have
   
   $$221k=18564$$
   
   Thus,
   
   $$k=\frac{18564}{221}=84$$
   
   So the correct answer is:
   
   $$\boxed{84}$$
   
   Hence, Option A is correct.
3. JEE Main 2026MediumSeries involving Products and Quotients of Binomial Coefficients
   Q3

   Let $S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\dots$ up to 13 terms. If $13S=\frac{2^k}{n!},k\in N$ , then $n+k$ is equal to

   • A50
   • B52
   • C49
   • D51

   Show answer & solution

   Correct answer: (C)

   $$S=\frac{1}{1!25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\dots 13\text{ terms }$$
   
   multiply in numerator and denominator by 26 !
   
   $$\begin{array}{rl}& S=\frac{1}{26!}\left[\begin{array}{r}\frac{26!}{1!25!}+\frac{26!}{3!23!}\\ +\frac{26!}{5!2!}+\dots 13\text{ term }\end{array}\right]\\ & S=\frac{1}{26!}\left[\begin{array}{r}{}^{26}{C}_1{+}^{26}{C}_3\\ {+}^{26}{C}_5+\dots 13\text{ term }\end{array}\right],\\ \left\{{}^n{C}_r=\frac{n!}{(n-r)!r!}\right\}& \end{array}$$
   
   Lets find 13th term lower subscript are in A.P with, common difference $=2$
   
   $$\begin{array}{rl}& T_{13}=1+(13-1)2=25\\ & \text{ so, }S=\frac{1}{26!}\left[{}^{26}{C}_1{+}^{26}{C}_3{+}^{26}{C}_5+{\dots }^{26}{C}_{25}\right]\\ & S=\frac{1}{26!}\left[2^{26-1}\right],\\ \left\{{\text{ because }}^n{C}_1{+}^n{C}_3{+}^n{C}_5\cdots =2^{n-1}\right\}& \end{array}$$
   
   $$S=\frac{2^{25}}{26!}$$
   
   It is given that $13S=\frac{2^K}{n!}$
   
   substitute value of $S$
   
   $$\begin{array}{rl}& 13\times \frac{2^{25}}{26!}=\frac{2^K}{n!}\\ & \frac{13\times 2^{25}}{26\times 25!}=\frac{2^K}{n!}\\ & \frac{2^{24}}{25!}=\frac{2^K}{n!}\end{array}$$
   
   so $n=25,K=24$
   
   $$n+K=25+24=49$$

Topic-wise question count

Where JEE Main has asked from Binomial Theorem, most frequent first.

• General Term and Middle Term(s) in Binomial Expansion57
• Properties and Sum of Single Binomial Coefficients45
• Term Independent of x & Rational/Irrational Terms33
• Divisibility, Remainder and Unit Digit Problems31
• Series involving Products and Quotients of Binomial Coefficients13
• Binomial Theorem for Any Index (Negative / Fractional)5
• Multinomial Theorem3
• Numerically Greatest Term in Binomial Expansion1

Year-wise Binomial Theorem questions in JEE Main

Number of questions from this chapter in each year's paper(s).