Binomial Theorem Previous Year Questions with Solutions

1.0Introduction 

Binomial Theorem Previous Year Questions typically cover topics like the general term, middle term, finding specific coefficients, and properties of binomial expansions. Examples include finding the term in an expansion, calculating the coefficient of a particular term in $(a+b{)}^n$, and using the binomial expansion to simplify expressions or solve identities. Solutions involve applying formulas such as the general term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$, understanding symmetry in coefficients, and identifying patterns. Practicing these questions helps in mastering expansion techniques and enhances algebraic manipulation skills critical for JEE-level problems.

2.0Binomial Theorem Previous Year Questions for JEE with Solutions

JEE questions in Binomial Theorem often test concepts related to binomial expansion, general term, middle term, and coefficients of specific terms. Some common types of problems include:

• General and Middle Term: Questions on finding the r^th term or the middle term in the expansion of expressions like $(x+a{)}^n.$
• Specific Coefficients: Problems that involve finding the coefficient of a particular term, such as x^r in the expansion of $(1+x{)}^n$ or $(a+bx{)}^n.$
• Greatest/Least Term and Term Independent of x: Questions involving identification of terms with maximum numerical value or the term independent of the variable.
• Properties and Identities: Applications of binomial identities, symmetry of binomial coefficients, and problems involving expansions with negative or fractional exponents (for JEE Advanced).

These questions are aimed at testing algebraic manipulation skills and a strong understanding of the binomial expansion formula.

Note: In the JEE Main Mathematics exam, you can generally expect 1 to 2 questions from the Binomial Theorem chapter.

3.0Key Concepts to Remember – Binomial Theorem (for JEE)

1. Binomial Expansion Formula:

$(a+b{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$

Where $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$ is the binomial coefficient.

2. General Term:

The $(r+1{)}^{th}$ term in the expansion of $(a+b{)}^n$ is given by:

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$

Used to find a specific term or coefficient in the expansion.

3. Middle Term(s):

• For even n: There is 1 middle term, at position $T_{\frac{n}{2}+1}$.
• For odd n: There are 2 middle terms, at positions $T_{\frac{n+1}{2}}\text{and}{T}_{\frac{n+3}{2}}$.

4. Term Independent of x:

To find the term independent of xx in an expression like $(ax+\frac{b}{x}{)}^n$, set the power of x in the general term to zero and solve for r.

5. Greatest Term:

Used when finding the numerically greatest term in an expansion for given values of x. Use approximation or comparison between consecutive terms.

6. Properties of Binomial Coefficients:

• Symmetry: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$
• Pascal’s Identity: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$
• Sum of all coefficients in $(a+b{)}^n:2^n(puta=b=1)$

7. Special Expansions:

• $(1+x{)}^n$ where n is a positive integer.
• $(1+x{)}^n$ where n is negative or fractional: Use binomial series expansion (for JEE Advanced only).

8. Applications in Inequalities and Approximations:

Used in approximation of expressions like $(1+x{)}^n$ for small x, or bounding polynomial values.

9. Negative and Fractional Index:

For |x| < 1,

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\dots$

Used when n is not a positive integer (JEE Advanced level).

Mastering these key concepts is essential to solving both direct and application-based problems from the Binomial Theorem in JEE.

4.0JEE Mains Past Year Questions with Solutions on Binomial Theorem 

1. If the coefficients of x⁴, x⁵ and x⁶ in the expansion of (1 + x)ⁿ are in the arithmetic progression, then the maximum value of n is : 

(1) 14

(2) 21

(3) 28

(4) 7

Ans. (1)

Sol.

Coeff. of x⁴ = ⁿC₄

Coeff. of x⁵ = ⁿC₅

Coeff. of x⁶ = ⁿC₆

ⁿC₄, ⁿC₅, ⁿC₆ …. AP

2.ⁿC₅ = ⁿC₄ + ⁿC₆

$$\begin{gathered} 2=\frac{{}^n{C}_4}{{}^n{C}_5}+\frac{{}^n{C}_6}{{}^n{C}_5}\left\{\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}\right\} \\ 2=\frac{5}{n-4}+\frac{n-5}{6} \end{gathered}$$

12(n – 4) = 30 + n² – 9n + 20

n² – 21n + 98 = 0

(n – 14) (n – 7) = 0

n_max = 14 n_min = 7

2. The sum of all rational terms in the expansion of \left( \frac{1}{2^{5}} + \frac{1}{5^3} \right)^{15}is equal to:

(1) 3133

(2) 633

(3) 931

(4) 6131

Ans. (1)

Sol.

$$\begin{gathered} T_{r+1}=15Cr{\left(\frac{1}{5^3}\right)}^r{\left(\frac{1}{25}\right)}^{15-r} \\ ={}^{15}{C}_r(\frac{r}{5^3}\cdot 2^{\frac{15-r}{5}} \end{gathered}$$

R = 3, 15µ

r = 0, 15

2 rational terms

${}^{15}{C}_0\cdot 2^3+{}^{15}{C}_{15}\cdot (5{)}^5$

= 8 + 3125 = 3133

3. Let $a=1+\frac{{}^2{C}_2}{3!}+\frac{{}^3{C}_2}{4!}+\frac{{}^4{C}_2}{5!}+\dots ,$

$b=1+\frac{{}^1{C}_0+{}^1{C}_1}{1!}+\frac{{}^2{C}_0+{}^2{C}_1+{}^2{C}_2}{2!}+\frac{{}^3{C}_0+{}^3{C}_1+{}^3{C}_2+{}^3{C}_3}{3!}+\dots$

Then $\frac{2b}{a^2}$ is equal to ______

Ans. (8)

Sol.

$$\begin{gathered} f(x)=1+\frac{(1+x)}{1!}+\frac{(1+x{)}^2}{2!}+\frac{(1+x{)}^3}{3!}+\dots \\ =\frac{e^{(1+x)}}{1+x}=1+\frac{1}{1!}+\frac{(1+x)}{2!}+\frac{(1+x{)}^2}{3!}+\frac{(1+x{)}^3}{4!}+\dots \\ \text{coef x2 in RHS :}1+\frac{{}^2{C}_2}{3!}+\frac{{}^3{C}_2}{4!}+\dots =a \\ \text{coeff. x2 in L.H.S. }=e\left(1+x+\frac{x^2}{2!}+\dots \right)\left(1-x+\frac{x^2}{2!}-\dots \right) \\ =e-e+\frac{e}{2!}=a \\ b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\dots =e^2 \\ \Rightarrow \frac{2b}{a^2}=8 \end{gathered}$$

4. If the constant term in the expansion of ${\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)}^{12}$, x ≠ 0, is α×2⁸×$\sqrt[5]{3}$

, then 25α is equal to :

(1) 639

(2) 724

(3) 693

(4) 742

Ans. (3)

Sol.

$$\begin{gathered} T_{r+1}={}^{12}{C}_r{\left(\frac{3^{1/2}}{x}\right)}^{12-r}{\left(\frac{2x}{5^{1/3}}\right)}^r \\ {T}_{r+1}={}^{12}{C}_r\cdot \frac{3^{(12-r)/2}\cdot 2^r\cdot x^r}{x^{12-r}\cdot 5^{r/3}}={}^{12}{C}_r\cdot \frac{3^{(12-r)/2}\cdot 2^r}{5^{r/3}}\cdot x^{2r-12} \\ \Rightarrow r=6 \\ {T}_7={}^{12}{C}_6\cdot \frac{3^{6/2}\cdot (2{)}^6}{5^2}=\frac{9\times 11\times 7}{25}\cdot 2^6\cdot 3^{1/5} \\ \Rightarrow 25a=693 \end{gathered}$$

5. If the constant term in the expansion of $(1+2x-3x^3){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$is p, then 108p is equal to 

Ans. (54)

Sol.  

$$\begin{gathered} (1+2x-3x^3){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9 \\ \text{General term m}{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9 \\ ={}^9{C}_r\cdot \frac{3^{9-r-r}}{2^{9-r}}\cdot (-1{)}^r\cdot x^{18-3r} \\ ={}^9{C}_r\cdot \frac{3^{9-2r}}{2^{9-r}}\cdot (-1{)}^r\cdot x^{18-3r} \\ \text{Put r = 6 to get coeff. of }{x}^0={}^9{C}_6\cdot \frac{1}{6^3}\cdot x^0=\frac{7}{18}{x}^0 \\ \text{Put r = 7 to get coeff. of }{x}^{-3}={}^9{C}_7\cdot \frac{3^5}{2^2}\cdot (-1{)}^7\cdot x^{-3} \\ =-{}^9{C}_7\cdot \frac{1}{3^5\cdot 2^2}\cdot x^{-3}=-\frac{1}{27}\cdot x^{-3} \\ (1+2x-3x^3)\left(\frac{7}{18}{x}^0-\frac{1}{27}{x}^{-3}\right) \\ \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} \\ 108\cdot \frac{1}{2}=54 \end{gathered}$$

6. If the second, third and fourth terms in the expansion of (x + y)ⁿ are 135, 30 and $\frac{10}{3}$, respectively, then $6(n^3+x^2+y)$ is equal to _____.

Ans. (806)

Sol.

$$\begin{gathered} {}^n{C}_1\cdot x^{n-1}\cdot y=135\dots \dots i \\ {}^n{C}_2\cdot x^{n-2}\cdot y^2=30\dots \dots (ii) \\ {}^n{C}_3\cdot x^{n-3}\cdot y^3=\frac{10}{3}\dots \dots (iii) \\ \frac{{}^n{C}_1\cdot x}{{}^n{C}_2\cdot y}=\frac{9}{2}\dots \dots (iv) \\ \frac{{}^n{C}_2\cdot x}{{}^n{C}_3\cdot y}=9\dots \dots (v) \\ \frac{{}^n{C}_1}{{}^n{C}_2}\cdot \frac{{}^n{C}_2}{{}^n{C}_3}=\frac{1}{2}\Rightarrow \frac{{}^n{C}_1}{{}^n{C}_3}=\frac{1}{2} \\ \Rightarrow \frac{n}{\frac{n(n-1)(n-2)}{6}}=\frac{1}{2} \\ \Rightarrow \frac{6n}{n(n-1)(n-2)}=\frac{1}{2} \\ \Rightarrow \frac{6}{(n-1)(n-2)}=\frac{1}{2} \\ \Rightarrow 12=(n-1)(n-2) \\ \Rightarrow n^2-3n+2=12 \\ \Rightarrow n^2-3n-10=0 \\ \Rightarrow (n-5)(n+2)=0 \\ \frac{x}{y}=9\Rightarrow x=9y \\ {}^5{C}_1\cdot x^4\cdot y=135 \\ \Rightarrow 5\cdot x^4\cdot y=135 \\ \Rightarrow x^4\cdot y=27 \\ \Rightarrow (9y{)}^4\cdot y=27 \\ \Rightarrow 6561y^5=27 \\ \Rightarrow y=\frac{1}{3},x=3 \\ 6(n^3+x^2+y) \\ 6(125+9+\frac{1}{3})=6\cdot \frac{406}{3}=\boxed{806} \end{gathered}$$

7. Let $\alpha ={\sum }_{r=0}^n(4r^2+2r+1)\cdot {}^n{C}_r,$ and $\beta =\left({\sum }_{r=0}^n\frac{{}^n{C}_r}{r+1}\right)+\frac{1}{n+1}$. If 140 < $\frac{2\alpha }{\beta }$ < 281, then the value of n is ______.

Ans. (5)

Sol.

$$\begin{gathered} \alpha ={\sum }_{r=0}^n(4r^2+2r+1)\cdot {}^n{C}_r \\ \alpha =4{\sum }_{r=0}^n{r}^2\cdot {}^n{C}_r+2{\sum }_{r=0}^{n}r\cdot {}^n{C}_r+{\sum }_{r=0}^n{}^n{C}_r \\ +4n{\sum }_{r=0}^{n}r\cdot {}^{n-1}{C}_{r-1}+2n{\sum }_{r=0}^n{}^{n-1}{C}_{r-1}+{\sum }_{r=0}^n{}^n{C}_r \\ \alpha =4n(n-1)\cdot 2^{n-2}+4n\cdot 2^{n-1}+2n\cdot 2^{n-1}+2^n \\ \alpha =2^{n-2}[4n(n-1)+8n+4n+4] \\ \alpha =2^{n-2}[4n^2+8n+4] \\ \alpha =2n(n+1{)}^2 \\ \beta ={\sum }_{r=0}^n\frac{{}^n{C}_r}{r+1}+\frac{1}{n+1} \\ ={\sum }_{r=0}^n\frac{{}^{n+1}{C}_{r+1}}{n+1}+\frac{1}{n+1} \\ =\frac{1}{n+1}\left({}^{n+1}{C}_1+{}^{n+1}{C}_2+\cdots +{}^{n+1}{C}_{n+1}\right) \\ =\frac{2^{n+1}}{n+1} \\ \frac{2\alpha }{\beta }=\frac{2n(n+1{)}^2}{\frac{2^{n+1}}{n+1}}=\frac{2^{n+1}(n+1{)}^2}{2^{n+1}}=(n+1{)}^3 \\ 140<(n+1{)}^3<281 \\ n=4\Rightarrow (n+1{)}^3=125 \\ n=5\Rightarrow (n+1{)}^3=216 \\ n=6\Rightarrow (n+1{)}^3=343 \\ \therefore n=5 \end{gathered}$$

8. The sum of the coefficient of x^2/3 and x^–2/5 in the binomial expansion of $\left(x^{⅔}+\frac{1}{2}{x}^{-⅖}\right)$

is :

(1) 21/4

(2) 69/16

(3) 63/16

(4) 19/4

Ans. (1)

Sol.

$$\begin{gathered} T_{r+1}={}^9{C}_r\cdot (x^{2/3}{)}^{9-r}\cdot {\left(\frac{x^{-2/5}}{2}\right)}^r \\ ={}^9{C}_r\cdot {\left(\frac{1}{2}\right)}^r\cdot x^{\left(\frac{2(9-r)}{3}-\frac{2r}{5}\right)} \\ Forthecoefficientof{x}^{2/3},put:\frac{2(9-r)}{3}-\frac{2r}{5}=\frac{2}{3} \\ \Rightarrow \frac{18-2r}{3}-\frac{2r}{5}=\frac{2}{3} \\ \Rightarrow 6-\frac{2r}{3}-\frac{2r}{5}=\frac{2}{3} \\ \Rightarrow -\frac{2r}{3}-\frac{2r}{5}=\frac{2}{3}-6=-\frac{16}{3} \\ \Rightarrow -2r\left(\frac{1}{3}+\frac{1}{5}\right)=-\frac{16}{3} \\ \Rightarrow -2r\cdot \frac{8}{15}=-\frac{16}{3} \\ \Rightarrow r=5 \\ \therefore \text{Coefficient of }{x}^{2/3}\text{ is }{}^9{C}_5{\left(\frac{1}{5}\right)}^5 \\ Nowforcoefficientof{x}^{-2/5}: \\ \frac{2(9-r)}{3}-\frac{2r}{5}=-\frac{2}{5} \\ \Rightarrow 6-\frac{2r}{3}-\frac{2r}{5}=-\frac{2}{5} \\ \Rightarrow -\frac{2r}{3}-\frac{2r}{5}=-\frac{2}{5}-6=-\frac{32}{5} \\ \Rightarrow -2r\left(\frac{1}{3}+\frac{1}{5}\right)=-\frac{32}{5} \\ \Rightarrow -2r\cdot \frac{8}{15}=-\frac{32}{5} \\ \Rightarrow r=6 \\ \text{Coefficient of }{x}^{-2/5}\text{ is }{}^9{C}_6\cdot {\left(\frac{1}{2}\right)}^6 \\ \text{Sum}={}^9{C}_5\cdot {\left(\frac{1}{2}\right)}^5+{}^9{C}_6\cdot {\left(\frac{1}{2}\right)}^6=\frac{21}{4} \end{gathered}$$

9. The remainder when 428²⁰²⁴ is divided by 21 is ________. Ans. (1)

Sol. (428)²⁰²⁴ = (420 + 8)²⁰²⁴

= (21 × 20 + 8)²⁰²⁴

= 21m  + 8²⁰²⁴

Now 8²⁰²⁴ = (8²)¹⁰¹²

= (64)¹⁰¹²

= (63 + 1)¹⁰¹²

= (21 × 3 + 1)¹⁰¹²

= 21n + 1

⇒ The remainder is 1.

10. If the Coefficient of x³⁰ in the expansion of ${\left(1+\frac{1}{x}\right)}^6$(1 + x²)⁷ (1 – x³)⁸ ; x ≠ 0 is α, then |α| equals ___________.

Ans. (678)

Sol.

$$\begin{gathered} \text{We need the coefficient of}{x}^{30}in: \\ \text{We need the coefficient of}{x}^{30}in: \\ \frac{(x+1{)}^6(1+x^2{)}^7(1-x^3{)}^8}{x^6} \\ \text{That is, the coefficient of}{x}^{36}in \\ (1+x{)}^6(1+x^2{)}^7(1-x^3{)}^8 \\ \text{Let us use the general term:} \\ {}^6{C}_{r_1}\cdot {}^7{C}_{r_2}\cdot {}^8{C}_{r_3}\cdot (-1{)}^{r_3}\cdot x^{r_1+2r_2+3r_3} \\ {r}_1+2r_2+3r_3=36 \end{gathered}$$$\text{Case I}\begin{array}{c}{r}_1{r}_2{r}_3\\ 068\\ 258\\ 448\\ 638\end{array}{r}_1+2r_2=12(\text{Taking}{r}_3=8)$

$\text{Case II}\begin{array}{c}{r}_1{r}_2{r}_3\\ 177\\ 367\\ 557\end{array}{r}_1+2r_2=15(\text{Taking}{r}_3=7)$

$\text{Case III}\begin{array}{c}{r}_1{r}_2{r}_3\\ 476\\ 666\end{array}{r}_1+2r_2=18(\text{Taking}{r}_3=6)$

$$\begin{gathered} =7+(15\cdot 21)+(15\cdot 35)+35 \\ -(6\cdot 8)-(20\cdot 7\cdot 8)-(6\cdot 21\cdot 8)+(15\cdot 28)+(7\cdot 28)=-678=\alpha \\ |\alpha |=678 \end{gathered}$$

11. The coefficient of x²⁰¹² in the expansion of $(1-x{)}^{2008}(1+x+x^2{)}^{2007}$ is equal to

Ans. (0)

Sol.

$$\begin{gathered} (1-x)(1-x^3{)}^{2007}(1+x+x^2{)}^{2007} \\ (1-x)(1-x^3{)}^{2007} \\ (1-x)\left(\left(\genfrac{}{}{0ex}{}{2007}{0}\right)-\left(\genfrac{}{}{0ex}{}{2007}{1}\right)x^3+\dots \right) \\ \text{General term} \\ (1-x)((-1{)}^r(\genfrac{}{}{0ex}{}{2007}{r})x^{3r}) \\ (-1{)}^r(\genfrac{}{}{0ex}{}{2007}{r})x^{3r}-(-1{)}^r\left(\genfrac{}{}{0ex}{}{2007}{r}\right)x^{3r+1} \\ 3r=2012 \\ \Rightarrow r=\frac{2012}{3}\text{(not an integer)} \\ 3r+1=2012 \\ \Rightarrow 3r=2011 \\ \Rightarrow r=\frac{2011}{3}\text{(not an integer)} \\ \text{Hence there is no term containing}{x}^{2012}. \\ \text{So coefficient of}{x}^{2012}=0 \end{gathered}$$

12. Remainder when $64^{32^{32}}$is divided by 9 is equal to ____.

Ans. 1

Sol.

$$\begin{gathered} Let32^{32}=t \\ 64^{32^{32}}=64^t=8^{2t}=(9-1{)}^{2t} \\ =9k+1 \\ \text{Hence remainder = 1 } \end{gathered}$$

13. Number of integral terms in the expansion of ${\left(\left(\genfrac{}{}{0ex}{}{1}{2}\right)x+\left(\genfrac{}{}{0ex}{}{1}{6}\right)\right)}^{824}$ is equal to ________.

Ans. (138)

Sol.

General term in expansion of ${\left((7{)}^{1/2}+(11{)}^{1/6}\right)}^{824}$ is 

t_{r + 1} = ⁸²⁴C_r

$t_{r+1}=\left(\genfrac{}{}{0ex}{}{824}{r}\right)(7{)}^{r/2}(11{)}^{(824-r)/6}$

For integral terms, r must be multiple of 6.

Hence r = 0, 6, 12, ……….822 

14. In the expansion of , $(1+x)(1-x^2){\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)}^5,x\ne 0$ the sum of the coefficient of x³ and x^–13 is equal to ___

Ans. (118)

Sol.

$$\begin{gathered} (1+x)(1-x^2){\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)}^5 \\ =(1+x)(1-x^2){\left({\left(1+\frac{1}{x}\right)}^3\right)}^5 \\ =\frac{(1+x{)}^2(1-x)(1+x{)}^{15}}{x^{15}} \\ =\frac{(1+x{)}^{17}-x(1+x{)}^{17}}{x^{15}} \\ =\text{coeff}(x^3)\text{ in the expansion}\approx \text{coeff}(x^{18})\text{ in }(1+x{)}^{17}-x(1+x{)}^{17} \\ =\left(\genfrac{}{}{0ex}{}{17}{18}\right)-\left(\genfrac{}{}{0ex}{}{17}{17}\right)=0-1=-1 \\ \text{coeff}(x^{-13})\text{ in the expansion}\approx \text{coeff}(x^2)\text{ in }(1+x{)}^{17}-x(1+x{)}^{17} \\ =\left(\genfrac{}{}{0ex}{}{17}{2}\right)-\left(\genfrac{}{}{0ex}{}{17}{1}\right)=136-17=119 \\ HenceAnswer=119-1=\boxed{118} \end{gathered}$$