NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem
Suppose you were asked to expand an algebraic expression like $(a+b)^{12}$. You can just multiply the expression 12 times over, but you will need to put in a lot of effort. In this scenario, the binomial theorem can make our lives much easier, providing us with a standard way to expand binomial expressions. The chapter Binomial Theorem in class 11 mathematics contains the concepts of binomial theorem for positive integral indices, Pascal’s triangle, and binomial theorem for any positive integer. Understanding these concepts will enhance the students' problem-solving ability and guide them in their advanced algebraic journey. The main objective of these Class 11 NCERT solutions is to prepare students efficiently and boost their confidence for the board exam and other competitive exams.
This Story also Contains
- NCERT Solution for Class 11 Maths Chapter 7 Solutions: Download PDF
- NCERT Solutions for Class 11 Maths Chapter 7: Exercise Questions
- Class 11 Maths NCERT Chapter 7: Extra Question
- Binomial Theorem Class 11 Chapter 7: Topics
- Binomial Theorem Class 11 Solutions: Important Formulae
- Approach to Solve Questions of Binomial Theorem Class 11
- What Extra Should Students Study Beyond NCERT for JEE?
- NCERT solutions for class 11 Maths: Chapter-Wise
This article on NCERT solutions for class 11 Maths Chapter 7 Binomial Theorem offers clear and step-by-step solutions for the exercise problems given in the NCERT book. Experienced Subject Matter Experts of Careers360 have curated these solutions following the latest NCERT syllabus, ensuring that students can grasp the basic concepts effectively. Many teachers rely on NCERT Solutions to explain concepts clearly to students. After going through these solutions, students can practice the NCERT exemplar and notes for more practice purposes.
NCERT Solution for Class 11 Maths Chapter 7 Solutions: Download PDF
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NCERT Solutions for Class 11 Maths Chapter 7: Exercise Questions
| Class 11 Maths chapter 7 solutions Exercise: 7.1 Page number: 132-133 Total questions: 14 |
Question 1: Expand the expression. $(1-2x)^5$
Answer:
Given,
The Expression:
$(1-2x)^5$
The expansion of this Expression is,
$(1-2x)^5 =$
$\\^5C_0(1)^5-^5C_1(1)^4(2x)+^5C_2(1)^3(2x)^2-^5C_3(1)^2(2x)^3+^5C_4(1)^1(2x)^4-^5C_5(2x)^5$
$=1-5(2x)+10(4x^2)-10(8x^3)+5(16x^4)-(32x^5)$
$=1-10x+40x^2-80x^3+80x^4-32x^5$
Question 2: Expand the expression. $\left(\frac{2}{x} - \frac{x}{2} \right )^5$
Answer:
Given,
The Expression:
$\left(\frac{2}{x} - \frac{x}{2} \right )^5$
the expansion of this Expression is,
$\left(\frac{2}{x} - \frac{x}{2} \right )^5=$
$\\^5C_0\left(\frac{2}{x}\right)^5-^5C_1\left(\frac{2}{x}\right)^4\left(\frac{x}{2}\right)+^5C_2\left(\frac{2}{x}\right)^3\left(\frac{x}{2}\right)^2$ $-^5C_3\left(\frac{2}{x}\right)^2\left(\frac{x}{2}\right)^3+^5C_4\left(\frac{2}{x}\right)^1\left(\frac{x}{2}\right)^4-^5C_5\left(\frac{x}{2}\right)^5$
$= \frac{32}{x}-5\left ( \frac{16}{x^4} \right )\left ( \frac{x}{2} \right )+10\left ( \frac{8}{x^3} \right )\left ( \frac{x^2}{4} \right )-10\left ( \frac{4}{x^2} \right )\left ( \frac{x^2}{8} \right )$ $+5\left ( \frac{2}{x} \right )\left ( \frac{x^4}{16} \right )-\frac{x^5}{32}$
$= \frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5x^2}{8}-\frac{x^3}{32}$
Question 3: Expand the expression. $(2x-3)^6$
Answer:
Given,
The Expression:
$(2x-3)^6$
The expansion of this Expression is,
$(2x-3)^6$
$= ^6C_0(2x)^6-^6C_1(2x)^5(3)+^6C_2(2x)^4(3)^2-^6C_3(2x)^3(3)^3+$ $^6C_4(2x)^2(3)^4-^6C_5(2x)(3)^5+^6C_6(3)^6$
$= 64x^6-6(32x^5)(3)+15(16x^4)(9)-20(8x^3)(27)+15(4x^2)(81)-6(2x)(243)$ $+729$
$= 64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729$
Question 4: Expand the expression. $\left(\frac{x}{3} + \frac{1}{x} \right )^5$
Answer:
Given,
The Expression:
$\left(\frac{x}{3} + \frac{1}{x} \right )^5$
The expansion of this Expression is,
$\left(\frac{x}{3} + \frac{1}{x} \right )^5$
$= ^5C_0\left(\frac{x}{3}\right)^5+^5C_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)+^5C_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2$ $+^5C_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3+^5C_4\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4+^5C_5\left(\frac{1}{x}\right)^5$
$= \frac{x^5}{243} +5\left ( \frac{x^4}{81} \right )\left ( \frac{1}{x} \right )+10\left ( \frac{x^3}{27} \right )\left ( \frac{1}{x^2} \right )+10\left ( \frac{x^2}{9} \right )\left ( \frac{1}{x^3} \right )$ $+5\left ( \frac{x}{3} \right )\left ( \frac{1}{x^4} \right )+\frac{1}{x^5}$
$= \frac{x^5}{243}+\frac{5x^3}{81}+\frac{10x}{27}+\frac{10}{9x}+\frac{5}{3x^2}+\frac{1}{x^5}$
Question 5: Expand the expression. $\left(x + \frac{1}{x} \right )^6$
Answer:
Given,
The Expression:
$\left(x + \frac{1}{x} \right )^6$
The expansion of this Expression is,
$\left(x + \frac{1}{x} \right )^6$
$= ^6C_0(x)^6+^6C_1(x)^5\left ( \frac{1}{x} \right )+^6C_2(x)^4\left ( \frac{1}{x} \right )^2+^6C_3(x)^3\left ( \frac{1}{x} \right )^3+$ $^6C_4(x)^2\left ( \frac{1}{x} \right )^4+^6C_5(x)\left ( \frac{1}{x} \right )^5+^6C_6\left ( \frac{1}{x} \right )^6$
$= x^6+6(x^5)\left ( \frac{1}{x} \right )+15(x^4)\left ( \frac{1}{x^2} \right )+20(8x^3)\left ( \frac{1}{x^3} \right )$ $+15(x^2)\left ( \frac{1}{x^4} \right )+6(x)\left ( \frac{1}{x^5} \right )+\frac{1}{x^6}$
$= x^6+6x^4+15x^2+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}$
Question 6: Using binomial theorem, evaluate the following: $(96)^3$
Answer:
As 96 can be written as (100-4);
$\\\Rightarrow (96)^3\\=(100-4)^3\\=^3C_0(100)^3-^3C_1(100)^2(4)+^3C_2(100)(4)^2-^3C_3(4)^3$
$=(100)^3-3(100)^2(4)+3(100)(4)^2-(4)^3$
$=1000000-120000+4800-64$
$=884736$
Question 7: Using binomial theorem, evaluate the following: $(102)^5$
Answer:
As we can write 102 in the form 100+2
$\Rightarrow (102)^5$
$\Rightarrow(100+2)^5$
$=^5C_0(100)^5+^5C_1(100)^4(2)+^5C_2(100)^3(2)^2$ $+^5C_3(100)^2(2)^3+^5C_4(100)^1(2)^4+^5C_5(2)^5$
$=10000000000+1000000000+40000000+800000+8000+32$
$=11040808032$
Question 8: Using binomial theorem, evaluate the following:
Answer:
As we can write 101 in the form 100 +1
$\Rightarrow (101)^4$
$\Rightarrow(100+1)^4$
$=^4C_0(100)^4+^4C_1(100)^3(1)+^4C_2(100)^2(1)^2$ $+^4C_3(100)^1(1)^3+^4C_4(1)^4$
$=100000000+4000000+60000+400+1$
$=104060401$
Question 9: Using binomial theorem, evaluate the following: $(99)^5$
Answer:
As we can write 99 in the form 100-1
$\Rightarrow (99)^5$
$\Rightarrow(100-1)^5$
$=^5C_0(100)^5-^5C_1(100)^4(1)+^5C_2(100)^3(1)^2$ $-^5C_3(100)^2(1)^3+^5C_4(100)^1(1)^4-^5C_5(1)^5$
$=10000000000-500000000+10000000-100000+500-1$
$= 9509900499$
Question 10: Using Binomial Theorem, indicate which number is larger (1.1) 10000 or 1000.
Answer:
AS we can write 1.1 as 1 + 0.1,
$(1.1)^{10000}=(1+0.1)^{10000}$
$=^{10000}C_0+^{10000}C_1(1.1)$+Other positive terms
$=1+10000\times1.1+$ Other positive term
$>1000$
Hence,
$(1.1)^{10000}>1000$
Question 11: Find $(a + b)^4 - (a-b)^4$ . Hence, evaluate $(\sqrt{3} + \sqrt2)^4 - (\sqrt3-\sqrt2)^4$.
Answer:
Using Binomial Theorem, the expressions $(a+b)^4$ and $(a-b)^4$ can be expressed as
$(a+b)^4=^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4$
$\Rightarrow (a-b)^4=^4C_0a^4-^4C_1a^3b+^4C_2a^2b^2-^4C_3ab^3+^4C_4b^4$
From Here,
$(a+b)^4-(a-b)^4 = ^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4$ $-^4C_0a^4+^4C_1a^3b-^4C_2a^2b^2+^4C_3ab^3-^4C_4b^4$
$\Rightarrow (a+b)^4-(a-b)^4 = 2\times( ^4C_1a^3b+^4C_3ab^3)$
$\Rightarrow (a+b)^4-(a-b)^4 = 8ab(a^2+b^2)$
Now, Using this, we get
$(\sqrt{3} + \sqrt2)^4 - (\sqrt3-\sqrt2)^4=8(\sqrt{3})(\sqrt{2})(3+2)=8\times\sqrt{6}\times5=40\sqrt{6}$
Question 12: Find $(x+1)^6 + (x-1)^6$ . Hence or otherwise evaluate $(\sqrt2+1)^6 + (\sqrt2-1)^6$.
Answer:
Using Binomial Theorem, the expressions $(x+1)^4$ and $(x-1)^4$ can be expressed as ,
$(x+1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+^6C_4x^21^4+^6C_5x1^5+^6C_61^6$
$\Rightarrow (x-1)^6=^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$
From Here,
$\\(x+1)^6-(x-1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+$ $^6C_4x^21^4+^6C_5x1^5+^6C_61^6$ $\:\:\:\:\;\:\:\;\:\:\:\ +^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$
$\Rightarrow (x+1)^6+(x-1)^6=2(^6C_0x^6+^6C_2x^41^2+^6C_4x^21^4+^6C_61^6)$
$\Rightarrow (x+1)^6+(x-1)^6=2(x^6+15x^4+15x^2+1)$
Now, Using this, we get
$(\sqrt2+1)^6 + (\sqrt2-1)^6=2((\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1)$
$\Rightarrow (\sqrt2+1)^6 + (\sqrt2-1)^6=2(8+60+30+1)=2(99)=198$
Question 13: Show that $9^{n+1} - 8n - 9$ is divisible by 64, whenever n is a positive integer.
Answer:
If we want to prove that $9^{n+1} - 8n - 9$ is divisible by 64, then we have to prove that $9^{n+1} - 8n - 9=64k$
As we know, from the binomial theorem,
$(1+x)^m=^mC_0+^mC_1x+^mC_2x^2+^mC_3x^3+....^mC_mx^m$
Here putting x = 8 and replacing m by n+1, we get,
$9^{n+1}=^{n+1}C_0+\:^{n+1}C_18+^{n+1}C_28^2+.......+^{n+1}C_{n+1}8^{n+1}$
$\Rightarrow 9^{n+1}=1+8(n+1)+8^2(^{n+1}C_2+\:^{n+1}C_38+^{n+1}C_48^2+.......+^{n+1}C_{n+1}8^{n-1})$
$\Rightarrow 9^{n+1}=1+8n+8+64(k)$
Now, Using This,
$9^{n+1} - 8n - 9=9+8n+64k-9-8n=64k$
Hence
$9^{n+1} - 8n - 9$ is divisible by 64.
Question 14: Prove that $\sum_{r = 0}^n3^r \ ^nC_r = 4^n$
Answer:
As we know from the Binomial Theorem,
$\sum_{r = 0}^na^r \ ^nC_r = (1+a)^n$
Here, putting a = 3, we get,
$\sum_{r = 0}^n3^r \ ^nC_r = (1+3)^n$
$\Rightarrow \sum_{r = 0}^n3^r \ ^nC_r = 4^n$
Hence Proved.
| Class 11 Maths chapter 7 solutions Miscellaneous exercise Page number: 133-133 Total questions: 6 |
Question 1: If a and b are distinct integers, prove that $a - b$ is a factor of $a^n - b^n$, whenever n is a positive integer.
[ Hint: write $a^n = (a - b + b)^n$ and expand]
Answer:
we need to prove,
$a^n-b^n=k(a-b)$ where k is some natural number.
Now let's add and subtract b from a so that we can prove the above result,
$a=a-b+b$
$a^n=(a-b+b)^n=[(a-b)+b]^n$
$=^nC_0(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_nb^n$
$=(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_{n-1}(a-b)b^{n-1}+b^n$ $\Rightarrow a^n-b^n=(a-b)[(a-b)^{n-1}+^nC_2(a-b)^{n-2}+........+^nC_{n-1}b^{n-1}]$
$\Rightarrow a^n-b^n=k(a-b)$
Hence, $a - b$ is a factor of $a^n - b^n$ .
Question 2: Evaluate $\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6$.
Answer:
First let's simplify the expression $(a+b)^6-(a-b)^6$ using binomial theorem,
So,
$(a+b)^6=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6b^6$
$\Rightarrow (a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$
And
$(a-b)^6=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6b^6$
$\Rightarrow (a+b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$
Now,
$(a+b)^6-(a-b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$ $-a^6+6a^5b-15a^4b^2+20a^3b^3-15a^2b^4+6ab^5-b^6$
$\Rightarrow (a+b)^6-(a-b)^6=2[6a^5b+20a^3b^3+6ab^5]$
Now, Putting $a=\sqrt{3}\:and\:b=\sqrt{2},$ we get
$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]$
$\Rightarrow \left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]$
$\Rightarrow \left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2\times198\sqrt{6}$
$\Rightarrow \left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=396\sqrt{6}$
Question 3: Find the value of $\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4$
Answer:
First, lets simplify the expression $(x+y)^4-(x-y)^4$ using binomial expansion,
$(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4$
$\Rightarrow (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$
And
$(x-y)^4=^4C_0x^4-^4C_1x^3y+^4C_2x^2y^2-^4C_3xy^3+^4C_4y^4$
$\Rightarrow (x-y)^4=x^4-4x^3y+6x^2y^2-4xy^3+y^4$
Now,
$(x+y)^4-(x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4-$ $x^4+4x^3y-6x^2y^2+4xy^3-y^4$
$\Rightarrow (x+y)^4-(x-y)^4=2(x^4+6x^2y^2+y^4)$
Now, Putting $x=a^2 and\:y=\sqrt{a^2-1}$ we get,
$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$
$\Rightarrow \left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[a^8+6a^4(a^2-1)+(a^2-1)^2]$
$\Rightarrow \left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-12a^4+2a^4-4a^2+2$
$\Rightarrow \left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-10a^4-4a^2+2$
Question 4: Find an approximation of (0.99) 5 using the first three terms of its expansion.
Answer:
As we can write 0.99 as 1-0.01,
$(0.99)^5=(1-0.001)^5=^5C_0(1)^5-^5C_1(1)^4(0.01)+^5C_2(1)^3(0.01)^2$ $+\:other \:negligible \:terms$
$\Rightarrow (0.99)^5=1-5(0.01)+10(0.01)^2$
$\Rightarrow (0.99)^5=1-0.05+0.001$
$\Rightarrow (0.99)^5=0.951$
Hence, the value of $(0.99)^5$ is approximately 0.951.
Question 5: Expand using Binomial Theorem $\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$
Answer:
Given the expression,
$\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$
The binomial expansion of this expression is
$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4\\=\left ( \left (1 + \frac{x}{2} \right ) -\frac{2}{x}\right )^4=^4C_0\left ( 1+\frac{x}{2} \right )^4-^4C_1\left ( 1+\frac{x}{2} \right )^3\left ( \frac{2}{x} \right )+$ $^4C_2\left ( 1+\frac{x}{2} \right )^2\left ( \frac{2}{x} \right )^2$ $-^4C_3\left ( 1+\frac{x}{2} \right )\left ( \frac{2}{x} \right )^3+^4C_4\left ( \frac{2}{x} \right )^4$
$= \left ( 1+\frac{x}{2} \right )^4-\frac{8}{x}\left ( 1+\frac{x}{2} \right )^3+$ $\frac{24}{x^2}+\frac{24}{x}+6$ $-\frac{32}{x^3}+\frac{16}{x^4}..........(1)$
Now applying the Binomial Theorem again,
$\left ( 1+\frac{x}{2} \right )^4=^4C_0(1)^4+^4C_1(1)^3\left ( \frac{x}{2} \right )+^4C_2(1)^2\left ( \frac{x}{2} \right )^2+^4C_3(1)\left ( \frac{x}{2} \right )^3$ $+^4C_4\left ( \frac{x}{2} \right )^4$
$= 1+ 4\left ( \frac{x}{2} \right )+6\left ( \frac{x^2}{4} \right )+4\left ( \frac{x^3}{8} \right )+\frac{x^4}{16}$
$=1+2x+\frac{3x^2}{2}+\frac{x^3}{3}+\frac{x^4}{16}..............(2)$
And
$\left ( 1+\frac{x}{2} \right )^3=^3C_0(1)^3+^3C_1(1)^2\left ( \frac{x}{2} \right )+^3C_2(1)\left ( \frac{x}{2} \right )^2+^3C_3\left ( \frac{x}{2} \right )^3$
$\Rightarrow \left ( 1+\frac{x}{2} \right )^3= 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}..........(3)$
Now, from (1), (2), and (3) we get,
$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$ $+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$
$\Rightarrow \\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2$ $+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$
$\Rightarrow \\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$ $=\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5$
Question 6: Find the expansion of $(3x^2 - 2ax +3a^2)^3$ using binomial theorem.
Answer:
Given $(3x^2 - 2ax +3a^2)^3$
By the Binomial Theorem, it can also be written as
$(3x^2 - 2ax +3a^2)^3=((3x^2 - 2ax) +3a^2)^3$
$=^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3$
$= (3x^2-2ax)^3+3(3x^2-2ax)^2(3a^2)+3(3x^2-2ax)(3a^2)^2+(3a^2)^3$
$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6$
$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6...........(1)$
Now, Again By Binomial Theorem,
$(3x^2-2ax)^3=^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3$
$\Rightarrow (3x^2-2ax)^3= 27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^2x^3$
$\Rightarrow (3x^2-2ax)^3= 27x^6-54x^5+36a^2x^4-8a^3x^3............(2)$
From (1) and (2), we get,
$(3x^2-2ax+3a^2)^3=27x^6-54x^5+36a^2x^3+81a^2x^4-108a^3x^3+117a^4x^2$ $-54a^5x+27a^6$
$\Rightarrow (3x^2-2ax+3a^2)^3=27x^6-54x^5+117a^2x^3-116a^3x^3+117a^4x^2$ $-54a^5x+27a^6$
Also Read,
Class 11 Maths NCERT Chapter 7: Extra Question
Question: If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression, then the maximum value of $n$ is:
Solution:
$\begin{aligned} & \text { Coeff. of } x^4={ }^n C_4 \\ & \text { Coeff. of } x^5={ }^n C_5 \\ & \text { Coeff. of } x^6={ }^n C_6 \\ & { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P \\ & 2 .{ }^n C_5={ }^n C_4+{ }^n C_6 \\ & ⇒2=\frac{{ }^n C_4}{{ }^n C_5}+\frac{{ }^n C_6}{{ }^n C_5} \quad\left\{\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{n-r+1}{r}\right\} \\ & ⇒2=\frac{5}{n-4}+\frac{n-5}{6} \\ & ⇒12(n-4)=30+n^2-9 n+20 \\ & ⇒n^2-21 n+98=0 \\ & ⇒(n-14)(n-7)=0 \\ & ⇒n_{\max }=14 \quad n_{\min }=7\end{aligned}$.
Therefore, the maximum value of $n$ is 14.
Binomial Theorem Class 11 Chapter 7: Topics
Given below are the topics discussed in the NCERT Solutions for class 11, chapter 7, Binomial Theorem:
- Introduction
- Binomial Theorem for Positive Integral Indices
Binomial Theorem Class 11 Solutions: Important Formulae
Binomial Theorem
The Binomial Theorem provides the expansion of a binomial (a + b) raised to any positive integer n.
The expansion of (a + b)n is given by: $(a+b)^n=^{n}\textrm{c}_0 a^n+^{n}\textrm{c}_1 a^{(n-1)} b+^{n}\textrm{c}_2 a^{(n-2)} b^2+\cdots+^{n}\textrm{c}_{n-1} \cdot a b^{(n-1)}+^{n}\textrm{c}_n b^n$
Special Cases from the Binomial Theorem:
$(x-y)^n=^{n}\textrm{c}_0 x^n-^{n}\textrm{c}_1 x^{(n-1)} y+^{n}\textrm{c}_2 x^{(n-2)} y^2+\ldots +(-1)^n . ^{n}\textrm{c}_n y^n$
$(1-x)^n=^{n}\textrm{c}_0-^{n}\textrm{c}_1 x+^{n}\textrm{c}_2n^2-\cdots+(-1)^n {^n c_n} x^n$
The coefficients $^{n}\textrm{c}_0$ and $^{n}\textrm{c}_n$ are always equal to 1.
Pascal’s Triangle
The coefficients of the expansions in the Binomial Theorem are arranged in an array called Pascal’s triangle.
General Terms of Expansion
For (a + b)n, the general term is $T_{r+1}=n_{C_r} a^{(n-r)} b^r$
For (a - b)n, the general term is $(-1)^r n c_r a^{(n-r)} b^r$
For (1 + x)n, the general term is $n_{C_r} x^r$
For (1 - x)n, the general term is $(-1)^r {^n c_n} x^n$
Middle Terms
In the expansion (a + b)n, if n is even, then the middle term is the ($\frac{n}{2}$ + 1)th term.
If n is odd, then the middle terms are the ($\frac{n}{2}$ + 1)th and ($\frac{n+1}{2}$ + 1)th terms.
Approach to Solve Questions of Binomial Theorem Class 11
Here are some steps on how to approach the questions of the Binomial Theorem:
- Learn the Basic Theorem: Before starting to solve any problem, you need to first understand the basic binomial theorem, i.e. $(a+b)^n=\sum\limits_{r=0}^n\binom{n}{r} a^{n-r} b^r$
Know what each term represents and where to apply the formula. - For a specific term related problem: If the question is asking for a specific term, then you can use the general term formula, i.e. $T_{r+1}=\binom{n}{r} a^{n-r} b^r$
- Coefficient-based problems: If you were asked to find the coefficient of a particular term, then write the general term and equate the exponent with the exponent of that term to find $r$, then calculate according to what is required.
- Mistakes to avoid: When you are trying to expand expressions like $(a-b)^n$, be careful about the alternate signs and remember that for even values of $r$ we will get positive terms and for odd values of $r$ we will get negative terms.
- Tips and tricks to improve speed and accuracy: To improve your speed and accuracy, you have to practice many different types of questions from the NCERT book, the exemplar book, and the previous year papers.
What Extra Should Students Study Beyond NCERT for JEE?
Here is a comparison list of the concepts in the Binomial Theorem that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:
NCERT solutions for class 11 Maths: Chapter-Wise
Given below is the chapter-wise list of the NCERT Class 11 Maths solutions with their respective links:
Also Read,
- NCERT Exemplar Class 11 Maths Solutions Chapter 7 Binomial Theorem
- NCERT Notes Class 11 Maths Chapter 7 Binomial Theorem
NCERT solutions for class 11- Subject-wise
Given below are some useful links for NCERT books and the NCERT syllabus for class 10:
- NCERT solutions for class 11 biology
- NCERT solutions for class 11 chemistry
- NCERT solutions for Class 11 physics
NCERT Books and NCERT Syllabus
Here are the subject-wise links for the NCERT solutions of class 10:
Frequently Asked Questions (FAQs)
Q: How many exercises are there in NCERT Class 11 Maths Chapter 7?
A:
There is one exercise in which binomial expansion-related questions are given and one miscellaneous exercise in which some advanced-level questions are given in NCERT Class 11 Maths Chapter 7 book.
Q: What is the general term in a binomial expansion?
A:
The general term of a binomial expansion $(a+b)^n$ is $T_{r+1} = nC_r a^{n-r} b^r$.
Q: What is the Binomial Theorem in NCERT Class 11 Maths?
A:
In class 11, the binomial theorem provides a formula to expand expressions of the form $(x + y)^n$, where 'n' is a positive integer, and 'x' and 'y' are any real numbers.
Expression is given by
$(x-y)^n=^{n}\textrm{c}_0 x^n-^{n}\textrm{c}_1 x^{(n-1)} y+^{n}\textrm{c}_2 x^{(n-2)} y^2+\ldots +(-1)^n . ^{n}\textrm{c}_n y^n$
Q: What is the middle term in binomial expansion?
A:
The middle term in a binomial expansion depends on whether the exponent 'n' is even or odd: if 'n' is even, there's one middle term, the $(\frac{n}{2} + 1)$ th term; if 'n' is odd, there are two middle terms, the $\frac{n+1}{2}$ and $\frac{n+3}{2}$ th terms.
Q: What are the applications of the Binomial Theorem in real life?
A:
The Binomial Theorem finds real-world applications in diverse fields like probability, statistics, economics, and engineering, allowing for calculations of probabilities in binomial distributions, economic forecasting, and estimating costs in construction projects.
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