Binomial Theorem JEE Main previous year questions with solutions

4 solved JEE Main questions on Binomial Theorem, free to read — no sign-in needed. The full chapter has 212 questions; sign in to attempt the remaining 208 in the exam simulator.

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1. Q1JEE Main 2026 (28 Jan, Shift 2)Integral and Fractional Part of a Number
   Given below are two statements : Statement I: $\quad 25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7. Statement II: The integral part of $(7+4 \sqrt{3})^{25}$ is an odd number. In the light of the above statements, choose the correct answer from the options given below :
   1. A.Statement I is true but Statement II is false
   2. B.Both Statement I and Statement II are true
   3. C.Statement I is false but Statement II is true
   4. D.Both Statement I and Statement II are false
   Show answer & solution

   Answer: (B)

   Statement I: $25^{13} + 3^{13}$ → divisible by $(25 + 3)$ $20^{13} + 8^{13}$ → divisible by $(20 + 8)$ $\therefore$ divisible by $7$. TRUE Statement II: For $(7+4\sqrt{3})^{25}$, let $\alpha = 7+4\sqrt{3}$ and $\beta = 7-4\sqrt{3}$ (conjugate). Then $\alpha\beta = 1$ and $\alpha + \beta = 14$. Since $0 \lt \beta \lt 1$, we have $\alpha^{25} + \beta^{25}$ is an integer. The fractional part of $\alpha^{25}$ is $1-\beta^{25}$, making the integral part odd. TRUE. Both Statement I and Statement II are true.
2. Q2JEE Main 2025 (03 Apr, Shift 2)Multinomial Theorem
   Let $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}$. If $\left(a_1+a_3+a_5+\ldots .+a_{19}\right)-11 \mathrm{a}_2=121 \mathrm{k}$, then k is equal to $\qquad$ .
   Show answer & solution

   Answer: 239

   $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}$ $\therefore 3^{10}=\mathrm{a}_0+\mathrm{a}_1+\mathrm{a}_2+\ldots .+\mathrm{a}_{20}$ ...(i) $1=a_0-a_1+a_2 \ldots . .+a_{20}$ ...(ii) $\text { (i) - (ii) } \Rightarrow a_1+a_3+\ldots .+a_{19}=\frac{3^{10}-1}{2}=29524$ $\begin{aligned} \text { Also }\{ & 1+\mathrm{x}(1+\mathrm{x})\}^{10}=1 \\ & +{ }^{10} \mathrm{C}_1 \mathrm{x}(1+\mathrm{x})+{ }^{10} \mathrm{C}_2 \mathrm{x}^2(1+\mathrm{x})^2+\ldots . \end{aligned}$ $\therefore \mathrm{a}_2={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2=55$ $\therefore \frac{\left(\mathrm{a}_1+\mathrm{a}_3+\ldots+\mathrm{a}_{19}\right)-11 \mathrm{a}_2}{121}=239$
3. Q3JEE Main 2024 (09 Apr, Shift 1)Remainder and Divisibility Problems
   The remainder when $428^{2024}$ is divided by 21 is__________
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   Answer: 1

   $\begin{aligned} & (428)^{2024}=(420+8)^{2024} \\ & =(21 \times 20+8)^{2024} \\ & =21 \mathrm{~m}+8^{2024} \end{aligned}$ Now $8^{2024}=\left(8^2\right)^{1012}$ $\begin{aligned} & =(64)^{1012} \\ & =(63+1)^{1012} \\ & =(21 \times 3+1)^{1012} \\ & =2 \ln +1 \end{aligned}$ $\Rightarrow$ Remainder is 1.
4. Q4JEE Main 2023 (10 Apr, Shift 2)Binomial Theorem for Negative Index
   If the coefficients of $x$ and ${x}^{2}$ in $(1+x{)}^{p}(1-x{)}^{q}$ are $4$ and $-5$ respectively, then $2p+3q$ is equal to
   1. A.$60$
   2. B.$69$
   3. C.$66$
   4. D.$63$
   Show answer & solution

   Answer: (D)

   Given that The coefficient of $x$ and ${x}^{2}$ in ${\left(1+x\right)}^{p}{\left(1-x\right)}^{q}$ are $4$ and $-5$. $\Rightarrow {\left(1+x\right)}^{p}{\left(1-x\right)}^{q}=\left(1+px+\dfrac{p\left(p-1\right)}{2!}{x}^{2}+....\right)\left(1-qx+\dfrac{q\left(q-1\right)}{2!}{x}^{2}-....\right)$ Now coefficient of $x$ from the above expansion will be $p-q$ which is equal to $4$ $\Rightarrow p-q=4$. Similarly coefficient of ${x}^{2}$ is $-5$. $\Rightarrow \dfrac{p\left(p-1\right)}{2}+\dfrac{q\left(q-1\right)}{2}-pq=-5$ $\Rightarrow \dfrac{{p}^{2}-2pq+{q}^{2}}{2}-\dfrac{\left(p+q\right)}{2}=-5$ $\Rightarrow \dfrac{{\left(p-q\right)}^{2}}{2}-\dfrac{\left(p+q\right)}{2}=-5$ $\Rightarrow \dfrac{16}{2}+5=\dfrac{\left(p+q\right)}{2}$ $\Rightarrow p+q=26$ and $p-q=4$ On solving the above equations we get, $p=15$ and $q=11$. $\Rightarrow 2p+3q=2\left(15\right)+3\left(11\right)=63$ Therefore, the required answer is $63$.