Chapter 7 Binomial Theorem

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NCERT Solutions Class 11 Maths Chapter 7 – Binomial Theorem

Class 11 Maths NCERT Solutions Chapter 7 ( Binomial Theorem ) In this chapter we learn a powerful tool of algebra which is used to expand expressions raised to any positive integral power. We know (a+b)^2 easily, and also (a+b)^3, but the Binomial Theorem allows us to compute (a+b)^{10} and more without tedious multiplication. This theorem is fundamental to algebra, calculus, and finite mathematics.

The NCERT Solutions for Class 11 Maths Chapter 7 from ALLEN are designed by expert faculty to simplify the process of expansion by providing clear explanations and a systematic approach toward the application of combinations (^nC_r). The solutions emphasise seeing the patterns in the exponents and coefficients, so that the formula is seen intuitively instead of being memorised as a string of variables.

Binomial Theorem is a must master for competitive exams where questions based on “independent terms” or “coefficient of x^n” are extremely common. These solutions offer a rigorous logical framework for students to engage with the properties of binomial coefficients and Pascal's Triangle.

1.0Class 11 Maths Chapter 7 Binomial Theorem: Key Concepts

This chapter focuses on the formula for expanding $(a + b)^{n}$ and the properties derived from it. Key lessons include:

Binomial Theorem for Positive Integral Indices: Understanding the general expansion formula:

$(a + b)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + \dots +^{n} C_{n} b^{n}$

Pascal’s Triangle: A geometric arrangement of binomial coefficients that helps in finding coefficients for small values of n.
General Term ( $T_{r + 1}$ ): A formula to find any specific term in the expansion without writing the full series:

$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$

Middle Term(s): Determining the central term in an expansion based on whether n is even or odd.
If n is even, there is one middle term: ( $\frac{n}{2} + 1)^{t h}$ term.
If n is odd, there are two middle terms: ( $\frac{n + 1}{2})^{t h} an d (\frac{n + 1}{2} + 1)^{t h}$ terms.
Term Independent of x: Learning how to find the constant term in an expansion where a and b contain powers of x.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 7

EXERCISE - 7.1

Expand each of the expressions in Q. 1 to 5.

1. $(1 - 2 x)^{5}$ Sol. By using Binomial Theorem, the expression $(1 - 2 x)^{5}$ can be expanded as

$(1 - 2 x)^{5} =^{5} C_{0} (1)^{5} -^{5} C_{1} (1)^{4} (2 x) +^{5} C_{2} (1)^{3} (2 x)^{2} -^{5} C_{3} (1)^{2} (2 x)^{3} +^{5} C_{4} (1)^{1} (2 x)^{4} -^{5} C_{5} (2 x)^{5} = 1 - 5 (2 x) + 10 (4 x^{2}) - 10 (8 x^{3}) + 5 (16 x^{4}) - (32 x^{5}) = 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}$

2. $(\frac{2}{x} - \frac{x}{2})^{5}$ Sol. By using Binomial Theorem, the expression $(\frac{2}{x} - \frac{x}{2})^{5}$ can be expanded as

$(\frac{2}{x} - \frac{x}{2})^{5} =^{5} C_{0} (\frac{2}{x})^{5} -^{5} C_{1} (\frac{2}{x})^{4} (\frac{x}{2}) +^{5} C_{2} (\frac{2}{x})^{3} (\frac{x}{2})^{2} -^{5} C_{3} (\frac{2}{x})^{2} (\frac{x}{2})^{3} +^{5} C_{4} (\frac{2}{x}) (\frac{x}{2})^{4} -^{5} C_{5} (\frac{x}{2})^{5} = \frac{32}{x ^{5}} - 5 (\frac{16}{x ^{4}}) (\frac{x}{2}) + 10 (\frac{8}{x ^{3}}) (\frac{x ^{2}}{4}) - 10 (\frac{4}{x ^{2}}) (\frac{x ^{3}}{8}) + 5 (\frac{2}{x}) (\frac{x ^{4}}{16}) - \frac{x ^{5}}{32} = \frac{32}{x ^{5}} - \frac{40}{x ^{3}} + \frac{20}{x} - 5 x + \frac{5}{8} x^{3} - \frac{x ^{5}}{32}$

3. $(2 x - 3)^{6}$ Sol. By using Binomial Theorem, the expression $(2 x - 3)^{6}$ can be expanded as

$(2 x - 3)^{6} =^{6} C_{0} (2 x)^{6} -^{6} C_{1} (2 x)^{5} (3) +^{6} C_{2} (2 x)^{4} (3)^{2} -^{6} C_{3} (2 x)^{3} (3)^{3} +^{6} C_{4} (2 x)^{2} (3)^{4} -^{6} C_{5} (2 x) (3)^{5} +^{6} C_{6} (3)^{6}$

$= 64 x^{6} - 6 (32 x^{5}) (3) + 15 (16 x^{4}) (9)$

$- 20 (8 x^{3}) (27) + 15 (4 x^{2}) (81) - 6 (2 x) (243) + 729 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} - 2916 x + 729$

4. $(\frac{x}{3} + \frac{1}{x})^{5}$ Sol. By using Binomial Theorem, the expression $(\frac{x}{3} + \frac{1}{x})^{5}$ can be expanded as

$(\frac{x}{3} + \frac{1}{x})^{5} =^{5} C_{0} (\frac{x}{3})^{5} +^{5} C_{1} (\frac{x}{3})^{4} (\frac{1}{x}) +^{5} C_{2} (\frac{x}{3})^{3} (\frac{1}{x})^{2} +^{5} C_{3} (\frac{x}{3})^{2} (\frac{1}{x})^{3} +^{5} C_{4} (\frac{x}{3}) (\frac{1}{x})^{4} +^{5} C_{5} (\frac{1}{x})^{5} = \frac{x ^{5}}{243} + 5 (\frac{x ^{4}}{81}) (\frac{1}{x}) + 10 (\frac{x ^{3}}{27}) (\frac{1}{x ^{2}}) + 10 (\frac{x ^{2}}{9}) (\frac{1}{x ^{3}}) + 5 (\frac{x}{3}) (\frac{1}{x ^{4}}) + \frac{1}{x ^{5}} = \frac{x ^{5}}{243} + \frac{5 x ^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x ^{3}} + \frac{1}{x ^{5}}$

5. $(x + \frac{1}{x})^{6}$ Sol. By using Binomial Theorem, the expression $(x + \frac{1}{x})^{6}$ can be expanded as

$(x + \frac{1}{x})^{6} =^{6} C_{0} (x)^{6} +^{6} C_{1} (x)^{5} (\frac{1}{x}) +^{6} C_{2} (x)^{4} (\frac{1}{x})^{2} +^{6} C_{3} (x)^{3} (\frac{1}{x})^{3} +^{6} C_{4} (x)^{2} (\frac{1}{x})^{4} +^{6} C_{5} (x) (\frac{1}{x})^{5} +^{6} C_{6} (\frac{1}{x})^{6}$

$= x^{6} + 6 (x)^{5} = x^{6} + 6 x^{4} (\frac{1}{x}) + 15 (x)^{4} (\frac{1}{x ^{2}}) + 20 (x)^{3} (\frac{1}{x ^{3}}) + 15 (x)^{2} (\frac{1}{x ^{4}}) + 6 (x) (\frac{1}{x ^{5}}) + \frac{1}{x ^{6}} + 15 x^{2} + 20 + \frac{15}{x ^{2}} + \frac{6}{x ^{4}} + \frac{1}{x ^{6}}$

Using Binomial Theorem, evaluate in Q. 6 to 9 6. $(96)^{3}$

Sol. It can be written that, $96 = 100 - 4$

$(96)^{3} = (100 - 4)^{3} =^{3} C_{0} (100)^{3} -^{3} C_{1} (100)^{2} (4) +^{3} C_{2} (100) (4)^{2} -^{3} C_{3} (4)^{3} = (100)^{3} - 3 (100)^{2} (4) + 3 (100) (4)^{2} - (4)^{3} = 1000000 - 120000 + 4800 - 64 = 884736$

7. $(102)^{5}$

Sol. It can be written that, $102 = 100 + 2$

$(102)^{5} = (100 + 2)^{5} =^{5} C_{0} (100)^{5} +^{5} C_{1} (100)^{4} (2) +^{5} C_{2} (100)^{3} (2)^{2} +^{5} C_{3} (100)^{2} (2)^{3} +^{5} C_{4} (100) (2)^{4} +^{5} C_{5} (2)^{5} = (100)^{5} + 5 (100)^{4} (2) + 10 (100)^{3} (2)^{2} + 10 (100)^{2} (2)^{3} + 5 (100) (2)^{4} + (2)^{5} = 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032$

8. $(101)^{4}$

Sol. It can be written that, $101 = 100 + 1$

$(101)^{4} = (100 + 1)^{4} =^{4} C_{0} (100)^{4} +^{4} C_{1} (100)^{3} (1) +^{4} C_{2} (100)^{2} (1)^{2} +^{4} C_{3} (100) (1)^{3} +^{4} C_{4} (1)^{4} = (100)^{4} + 4 (100)^{3} + 6 (100)^{2} + 4 (100) + (1)^{4} = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401$

9. $(99)^{5}$

Sol. It can be written that, $99 = 100 - 1$

$(99)^{5} = (100 - 1)^{5} =^{5} C_{0} (100)^{5} -^{5} C_{1} (100)^{4} (1) +^{5} C_{2} (100)^{3} (1)^{2} -^{5} C_{3} (100)^{2} (1)^{3} +^{5} C_{4} (100) (1)^{4} -^{5} C_{5} (1)^{5} = (100)^{5} - 5 (100)^{4} + 10 (100)^{3} - 10 (100)^{2} + 5 (100) - 1 = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 = 10010000500 - 500100001 = 9509900499$

10.Using Binomial Theorem, indicate which number is larger (1.1) $^{10000}$ or 1000.

Sol. By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1) $^{10000}$ can be obtained as

$(1.1)^{10000} = (1 + 0.1)^{10000} =^{10000} C_{0} +^{10000} C_{1} (1.1) + other positive terms = 1 + 10000 \times 1.1 + other positive terms = 1 + 11000 + other positive terms > 1000$

Hence, (1.1) $^{10000} > 1000$ 11. Find $(a + b)^{4} - (a - b)^{4}$ . Hence, evaluate

$(3 + 2)^{4} - (3 - 2)^{4}$

Sol. Using Binomial Theorem, the expressions, $(a + b)^{4}$ and $(a - b)^{4}$ , can be expanded as

$(a + b)^{4} =^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4} (a - b)^{4} =^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4}$

$∴ - (a + b)^{4} - (a - b)^{4} =^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4} [^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4}]$

$= 2 (^{4} C_{1} a^{3} b +^{4} C_{3} a b^{3}) = 2 (4 a^{3} b + 4 a b^{3}) = 8 ab (a^{2} + b^{2})$

By putting $a = 3$ and $b = 2$ , we obtain

$(3 + 2)^{4} - (3 - 2)^{4} = 8 (3) (2) {(3)^{2} + (2)^{2}} = 8 (6) (3 + 2) = 406$

12.Find $(x + 1)^{6} + (x - 1)^{6}$ . Hence or otherwise evaluate. $(2 + 1)^{6} + (2 - 1)^{6}$

Sol. Using Binomial Theorem, the expressions, (x $+ 1)^{6}$ and $(x - 1)^{6}$ , can be expanded as

$(x + 1)^{6} =^{6} C_{0} x^{6} +^{6} C_{1} x^{5} +^{6} C_{2} x^{4} +^{6} C_{3} x^{3} +^{6} C_{4} x^{2} +^{6} C_{5} x +^{6} C_{6} (x - 1)^{6} =^{6} C_{0} x^{6} -^{6} C_{1} x^{5} +^{6} C_{2} x^{4} -^{6} C_{3} x^{3} -^{6} C_{4} x^{2} +^{6} C_{5} x +^{6} C_{6}$

$∴ (x + 1)^{6} + (x - 1)^{6}$

$= 2 [^{6} C_{0} x^{6} +^{6} C_{2} x^{4} +^{6} C_{4} x^{2} +^{6} C_{6}] = 2 [x^{6} + 15 x^{4} + 15 x^{2} + 1]$

By putting $x = 2$ we obtain

$= = = = = (2 + 1)^{6} + (2 - 1)^{6} [(2)^{6} + 15 (2)^{4} + 15 (2)^{2} + 1] 2 (8 + 15 \times 4 + 15 \times 2 + 1) 2 (8 + 60 + 30 + 1) 2 (99) 198$

13.Show that $9^{n + 1} - 8 n - 9$ is divisible by 64, whenever n is a positive integer.

Sol. In order to show that $9^{n + 1} - 8 n - 9$ is divisible by 64, it has to be proved that $9^{n + 1} - 8 n - 9 = 64 k$ , where k is some natural number.By Binomial Theorem ,

$(1 + a)^{m} =^{m} C_{0} +^{m} C_{1} a +^{m} C_{2} a^{2} + \dots +^{m} C_{m} a^{m}$

For $a = 8$ and $m = n + 1$ , we obtain

$(1 + 8)^{n + 1} =^{n + 1} C_{0} +^{n + 1} C_{1} (8) +^{n + 1} C_{2} (8)^{2} + \dots .. +^{n + 1} C_{n + 1} (8)^{n + 1}$

$\Rightarrow 9^{n + 1} = 1 + (n + 1) (8)$

$+ (8)^{2} [^{n + 1} C_{2} +^{n + 1} C_{3} \times 8 + \dots +^{n + 1} C_{n + 1} (8)^{n - 1}]$

$\Rightarrow 9^{n + 1} = 9 + 8 n$

$+ 64 [^{n + 1} C_{2} +^{n + 1} C_{3} \times 8 + \dots +^{n + 1} C_{n + 1} (8)^{n - 1}]$

$\Rightarrow 9^{n + 1} - 8 n - 9 = 64 k$ , where

$k =^{n + 1} C_{2} +^{n + 1} C_{3} \times 8 + \dots .. +^{n + 1} C_{n + 1} (8)^{n - 1} is a$

natural number. Thus, $9^{n + 1} - 8 n - 9$ is divisible by 64, whenever n is a positive interger. 14. Prove that : $\sum_{r = 0}^{n} 3^{r}^{n} C_{r} = 4^{n}$

Sol. By Binomial Theorem,

$\sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} = (a + b)^{n}$

By putting $b = 3$ and $a = 1$ in the above equation, we obtain

$\sum_{r = 0}^{n}^{n} C_{r} (1)^{n - r} (3)^{r} = (1 + 3)^{n}$

$\Rightarrow \sum_{r = 0}^{n} 3^{r} C_{r} = 4^{n}$ Hence Proved.

3.0MISCELLANEOUS EXERCISE

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]

Sol. In order to prove that $(a - b)$ is a factor of $(a^{n} -$ $b^{n}$ ), it has to be proved that $a^{n} - b^{n} = k (a - b)$ , where k is some natural number. It can be written that, $a = a - b + b$

$∴ a^{n} = (a - b + b)^{n} = [(a - b) + b]^{n}$

$=^{n} C_{0} (a - b)^{n} +^{n} C_{1} (a - b)^{n - 1} b + \dots\dots +^{n} C_{n - 1} (a - b) b^{n - 1} +^{n} C_{n} b^{n} = (a - b)^{n} +^{n} C_{1} (a - b)^{n - 1} b + \dots\dots . +^{n} C_{n - 1} (a - b) b^{n - 1} + b^{n}$

$\Rightarrow a^{n} - b^{n} = (a - b)$

$[(a - b)^{n - 1} +^{n} C_{1} (a - b)^{n - 2} b + \dots . +^{n} C_{n - 1} b^{n - 1}]$

$\Rightarrow a^{n} - b^{n} = k (a - b)$

where,

$k = [(a - b)^{n - 1} +^{n} C_{1} (a - b)^{n - 2} b + \dots . +^{n} C_{n - 1} b^{n - 1}]$

is a natural number. This shows that $(a - b)$ is a factor of $(a^{n} - b^{n})$ , where n is a positive integer.

2.Evaluate : $(3 + 2)^{6} - (3 - 2)^{6}$

Sol. Firstly, the expression $(a + b)^{6} - (a - b)^{6}$ is simplified by using Binomial Theorem. This can be done as

$(a + b)^{6} =^{6} C_{0} a^{6} +^{6} C_{1} a^{5} b +^{6} C_{2} a^{4} b^{2} +^{6} C_{3} a^{3} b^{3} +^{6} C_{4} a^{2} b^{4} +^{6} C_{5} a^{1} b^{5} +^{6} C_{6} b^{6} = a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}$

$∴ (a + b)^{6} - (a - b)^{6} = 2 [6 a^{5} b + 20 a^{3} b^{3} + 6 a b^{5}]$ Putting $a = 3$ and $b = 2$ , we obtain

$(3 + 2)^{6} - (3 - 2)^{6}$

3.Find the value of $(a^{2} + a^{2} - 1)^{4} + (a^{2} - a^{2} - 1)^{4}$

Sol. Firstly, the expression $(x + y)^{4} + (x - y)^{4}$ is simplified by using Binomial Theorem. This can be done as

$(x + y)^{4} = =^{4} C_{0} x^{4} +^{4} C_{1} x^{3} y +^{4} C_{2} x^{2} y^{2} +^{4} C_{3} x y^{3} +^{4} C_{4} y^{4} x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} (x - y)^{4} =^{4} C_{0} x^{4} -^{4} C_{1} x^{3} y +^{4} C_{2} x^{2} y^{2} - -^{4} C_{3} x y^{3} +^{4} C_{4} y^{4} = x^{4} - 4 x^{3} y + 6 x^{2} y^{2} - 4 x y^{3} + y^{4} ∴ (x + y)^{4} + (x - y)^{4} = 2 (x^{4} + 6 x^{2} y^{2} + y^{4}) Putting x = a^{2} and y = a^{2} - 1, we obtain (a^{2} + a^{2} - 1)^{4} + (a^{2} - a^{2} - 1)^{4} = 2 [(a^{2})^{4} + 6 (a^{2})^{2} (a^{2} - 1)^{2} + (a^{2} - 1)^{4}] = 2 [a^{8} + 6 a^{4} (a^{2} - 1) + (a^{2} - 1)^{2}] = 2 [a^{8} + 6 a^{6} - 6 a^{4} + a^{4} - 2 a^{2} + 1] = 2 a^{8} + 12 a^{6} - 10 a^{4} - 4 a^{2} + 2$

4.Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.

Sol. $0.99 = 1 - 0.01$

$∴ = = = = = (0.99)^{5} = (1 - 0.01)^{5}^{5} C_{0} (1)^{5} -^{5} C_{1} (1)^{4} (0.01) +^{5} C_{2} (1)^{3} (0.01)^{2} -^{5} C_{3} (1)^{4} (0.01)^{3} + \dots\dots (Approximately) 1 - 5 (0.01) + 10 (0.01)^{2} 1 - 0.05 + 0.001 1.001 - 0.050 0.951$

Thus, the value of $(0.99)^{5}$ is approximately 0.951.

5.Expand using Binomial Theorem

$(1 + \frac{x}{2} - \frac{2}{x})^{4}, x \neq = 0$

Sol. Using Binomial Theorem, the given expression $(1 + \frac{x}{2} - \frac{2}{x})^{4}$ can be expanded as

$[(1 + \frac{x}{2}) - \frac{2}{x}]^{4} =^{4} C_{0} (1 + \frac{x}{2})^{4} -^{4} C_{1} (1 + \frac{x}{2})^{3} (\frac{2}{x}) +^{4} C_{2} (1 + \frac{x}{2})^{2} (\frac{2}{x})^{2} -^{4} C_{3} (1 + \frac{x}{2}) (\frac{2}{x})^{3} +^{4} C_{4} (\frac{2}{x})^{4} = (1 + \frac{x}{2})^{4} - 4 (1 + \frac{x}{2})^{3} (\frac{2}{x}) + 6 (1 + x + \frac{x ^{2}}{4}) (\frac{4}{x ^{2}}) - 4 (1 + \frac{x}{2}) (\frac{8}{x ^{3}}) + \frac{16}{x ^{4}} = (1 + \frac{x}{2})^{4} - \frac{8}{x} (1 + \frac{x}{2})^{3} + \frac{24}{x ^{2}} + \frac{24}{x} + 6 - \frac{32}{x ^{3}} - \frac{16}{x ^{2}} + \frac{16}{x ^{4}} = (1 + \frac{x}{2})^{4} - \frac{8}{x} (1 + \frac{x}{2})^{3} + \frac{8}{x ^{2}} + \frac{24}{x} + 6 - \frac{32}{x ^{3}} + \frac{16}{x ^{4}}$

Again by using Binomial Theorem, we obtain

$(1 + \frac{x}{2})^{4} =^{4} C_{0} (1)^{4} -^{4} C_{1} (1)^{3} (\frac{x}{2}) +^{4} C_{2} (1)^{2} (\frac{x}{2})^{2} -^{4} C_{3} (1)^{1} (\frac{x}{2})^{3} +^{4} C_{4} (\frac{x}{2})^{4} = 1 + 4 \times \frac{x}{2} + 6 \times \frac{x ^{2}}{4} + 4 \times \frac{x ^{3}}{8} + \frac{x ^{4}}{16} = 1 + 2 x + \frac{3 x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16}$

$(1 + \frac{x}{2})^{3} = = 1 + \frac{3 x}{2} + \frac{3 x ^{2}}{4} + \frac{x ^{3}}{8}^{3} C_{0} (1)^{3} -^{3} C_{1} (1)^{2} (\frac{x}{2}) +^{3} C_{2} (1) (\frac{x}{2})^{2} -^{3} C_{3} (\frac{x}{2})^{3} \dots\dots .$

From (1), (2) and (3) we obtain

$[(1 + \frac{x}{2}) - \frac{2}{x}]^{4} = 1 + 2 x + \frac{3 x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16} - \frac{8}{x} (1 + \frac{3 x}{2} + \frac{3 x ^{2}}{4} + \frac{x ^{3}}{8}) + \frac{8}{x ^{2}} + \frac{24}{x} + 6 - \frac{32}{x ^{3}} + \frac{16}{x ^{4}} = 1 + 2 x + \frac{3}{2} x^{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16} - \frac{8}{x} - 12 - 6 x - x^{2} + \frac{8}{x ^{2}} + \frac{24}{x} + 6 - \frac{32}{x ^{3}} + \frac{16}{x ^{4}} = \frac{16}{x} + \frac{8}{x ^{2}} - \frac{32}{x ^{3}} + \frac{16}{x ^{4}} - 4 x + \frac{x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16} - 5$

6.Find the expansion of $(3 x^{2} - 2 a x + 3 a^{2})^{3}$ using binomial theorem.

Sol. Using Binomial Theorem, the given expression $(3 x^{2} - 2 a x + 3 a^{2})^{3}$ can be expanded as

$= = = = [(3 x^{2} - 2 a x) + 3 a^{2}]^{3}^{3} C_{0} (3 x^{2} - 2 a x)^{3} +^{3} C_{1} (3 x^{2} - 2 a x)^{2} (3 a^{2}) +^{3} C_{2} (3 x^{2} - 2 a x) (3 a^{2})^{2} +^{3} C_{3} (3 a^{2})^{3} (3 x^{2} - 2 a x)^{3} + 3 (9 x^{4} - 12 a x^{3} + 4 a^{2} x^{2}) (3 a^{2}) + 3 (3 x^{2} - 2 a x) (9 a^{4}) + 27 a^{6} (3 x^{2} - 2 a x)^{3} + 81 a^{2} x^{4} - 108 a^{3} x^{3} + 36 a^{4} x^{2} + 81 a^{4} x^{2} - 54 a^{5} x + 27 a^{6} (3 x^{2} - 2 a x)^{3} + 81 a^{2} x^{4} - 108 a^{3} x^{3} + 117 a^{4} x^{2} - 54 a^{5} x + 27 a^{6}$

Again by using Binomial Theorem, we obtain

$(3 x^{2} - 2 a x)^{3} =^{3} C_{0} (3 x^{2})^{3} -^{3} C_{1} (3 x^{2})^{2} (2 a x) +^{3} C_{2} (3 x^{2}) (2 a x)^{2} +^{3} C_{3} (2 a x)^{3} = 27 x^{6} - 3 (9 x^{4}) (2 a x) + 3 (3 x^{2}) (4 a^{2} x^{2}) - 8 a^{3} x^{3} = 27 x^{6} - 54 a x^{5} + 36 a^{2} x^{4} - 8 a^{3} x^{3}$

From (1) and (2), we obtain

$(3 x^{2} - 2 a x + 3 a^{2})^{3} = 27 x^{6} - 54 a x^{5} + 36 a^{2} x^{4} - 8 a^{3} x^{3} + 81 a^{2} x^{4} - 108 a^{3} x^{3} + 117 a^{4} x^{2} - 54 a^{5} x + 27 a^{6} = 27 x^{6} - 54 a x^{5} + 117 a^{2} x^{4} - 116 a^{3} x^{3} + 117 a^{4} x^{2} - 54 a^{5} x + 27 a^{6}$

4.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Get complete NCERT Solutions for Class 11 Maths, organised by chapter with exercise answers, useful formulas, and clear steps to understand and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

5.0NCERT Class 11 Maths Chapter 7: Binomial Theorem Exercise-wise Questions and Topics

Exercise	Number of Questions	Important Topics Covered
Exercise 7.1	14 Questions and Solutions	Binomial expansion, middle terms, and applications of the Binomial Theorem
Miscellaneous Exercise	6 Questions and Solutions	Number of terms, sum of terms, and applications of the Binomial Theorem

6.0Key Features of NCERT Solutions for Class 11 Maths Chapter 7 (Binomial Theorem)

Clear Understanding of Binomial Expansion: Learn the pattern of a binomial expansion, where the power of the first term decreases while the power of the second term increases in each successive term.
Simplified General Term Application: Expert guidance on solving "Find the coefficient of $x^{k}$ " problems by setting the exponent in the general term equal to k and solving for r.
Middle Term Logic: Learn how to identify the middle term or middle terms of a binomial expansion based on whether the power of the binomial is even or odd
Efficient Calculation of Coefficients: The solutions provide tips on using properties of $^{n} C_{r} (l ik e^{n} C_{r} =^{n} C_{n - r})$ to speed up calculations during exams.
Step-by-Step Solutions: Each question is explained through clear mathematical steps, making it easier to understand binomial expansion, coefficients, and term-related questions.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with clear solutions to help students understand and apply the Binomial Theorem.

Frequently Asked Questions

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Frequently Asked Questions

Why are NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem important for students?

How do Class 11 Maths NCERT Solutions for Chapter 7 help in board and JEE preparation?

What concepts are covered in NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem?

Are NCERT Solutions for Class 11 Maths Chapter 7 by ALLEN useful for competitive exams?

Why is the Binomial Theorem important in mathematics?

How do you find the middle term in a binomial expansion?

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NCERT Solutions Class 11 Maths Chapter 7 – Binomial Theorem

1.0Class 11 Maths Chapter 7 Binomial Theorem: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 7

3.0MISCELLANEOUS EXERCISE

4.0Class 11 Maths NCERT Solutions – Chapter-wise Links

5.0NCERT Class 11 Maths Chapter 7: Binomial Theorem Exercise-wise Questions and Topics

6.0Key Features of NCERT Solutions for Class 11 Maths Chapter 7 (Binomial Theorem)

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