Class 11 Mathematics Chapter 7 Binomial Theorem

Key Concept: Number of Terms

d) 8

[Solution Description] According to the properties of binomial expansions, the number of terms in the expansion of $(a+b)^n$ is $n+1$. Here, with $n = 7$, the number of terms is $7 + 1 = 8$.

Key Concept: Identify Coefficients

d) 54

[Solution Description] To find the coefficient of $x^2$ in the expansion of $(x + 3)^4$, we use the binomial theorem:

$$(x + 3)^4 = \sum_{r=0}^{4} \binom{4}{r} x^{4-r} \cdot 3^r$$

We are looking for the coefficient of the term where $x^2$ appears, which means $4 – r = 2$. Hence, $r = 2$. The term corresponding to $r = 2$ is:

$$\binom{4}{2} x^{4-2} \cdot 3^2 = \binom{4}{2} x^2 \cdot 9$$

Calculating $$\binom{4}{2}$$:

$$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$$

Thus, the coefficient is $$6 \cdot 9 = 54$$.

Key Concept: Coefficient Identification

c) 6

[Solution Description] In the expansion of $(x + y)^4$, we use the binomial theorem:

$$\binom{4}{k} x^{4-k} y^k$$

We need the term where $x^2y^2$. This happens when $4-k = 2$ and $k = 2$. So:

$$\binom{4}{2} x^2 y^2 = \frac{4!}{2!2!} \cdot x^2 y^2 = \frac{24}{4} \cdot x^2 y^2 = 6x^2 y^2$$

The coefficient of $x^2y^2$ is 6.

Key Concept: Binomial Coefficients

b) 35

[Solution Description] The binomial coefficient for a term in the expansion of $(x+y)^n$ is given by $\binom{n}{k}$ where $k$ is the power of $x$. Here, $n = 7$ and the power of $x$, which is $k = 3$. So, we calculate:

$$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$

Thus, the binomial coefficient is 35.

Key Concept: Number of Terms

b) 7

[Solution Description] The number of terms in the expansion of a binomial expression $(a + b)^n$ is $n + 1$. Therefore, for $(a + b)^6$, the number of terms is $6 + 1 = 7$.

Key Concept: Pascal’s Triangle Basics

c) 10

[Solution Description] To find the fourth number in the row corresponding to index 5 in Pascal’s Triangle, we identify it as $^5C_3$. Using the combination formula:

$$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$

Thus, the fourth number is 10.

Key Concept: Coefficient Identification

c) 1, 4, 6, 4, 1

[Solution Description] To find the coefficients for $(a + b)^4$, we look at the fifth row of Pascal’s Triangle (since we start counting from the top row as 0):

The fifth row is: 1, 4, 6, 4, 1

Therefore, the coefficients are 1, 4, 6, 4, 1.

Key Concept: Basic Expansion

b) $x^2 + 2x + 1$

[Solution Description] The binomial expansion for a small power like 2 can be found directly using the formula $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = x$ and $b = 1$. Therefore, the expansion is:

– Calculate $a^2$: $$(x)^2 = x^2$$

– Calculate $2ab$: $$2 \cdot x \cdot 1 = 2x$$

– Calculate $b^2$: $$(1)^2 = 1$$

Combine these results to get $$x^2 + 2x + 1$$.

Key Concept: Pattern Recognition

a) 1, 7, 21, 35, 35, 21, 7, 1

[Solution Description] The given row is 1, 6, 15, 20, 15, 6, 1 which corresponds to the expansion coefficients of $(a + b)^6$. Each element is calculated as the sum of the two elements above it. Therefore, the next row will correspond to $(a + b)^7$ and will be formed by summing adjacent values:1,

$1+6 = 7$,

$6+15 = 21$,

$15+20 = 35$,

$20+15 = 35$,

$15+6 = 21$,

$6+1 = 7$,

1.

Thus, the next row is 1, 7, 21, 35, 35, 21, 7, 1.

Key Concept: Sum of Coefficients

d) 32

[Solution Description] The sum of the coefficients in the expansion of $$(x+1)^n$$ is calculated using the formula $2^n$. For $$(x+1)^5$$, we compute:

$$2^5 = 32$$

Therefore, the sum of the coefficients is 32.

Key Concept: Basic Expansion

b) $x^2 – 2xy + y^2$

[Solution Description] The binomial expansion for $(x-y)^n$ is given by:

$$(x-y)^2 = nC0 \cdot x^2 – nC1 \cdot x^{2-1}y + nC2 \cdot y^2$$

Here, $n = 2$. Calculating each term:

– First term: $$2C0 \cdot x^2 = 1 \cdot x^2 = x^2$$

– Second term: $$2C1 \cdot x^1 \cdot y = 2 \cdot xy = -2xy$$

– Third term: $$2C2 \cdot y^2 = 1 \cdot y^2 = y^2$$

Thus, the expansion is: $x^2 – 2xy + y^2$

Key Concept: Basic Expansion

a) $x^2 – 2xy + y^2$

[Solution Description] Using the expansion formula for $(x-y)^n$, we have:

$$(x – y)^2 = 2C0 \cdot x^2 – 2C1 \cdot x^{1} \cdot y + 2C2 \cdot y^2$$

Calculating each term, we get:

For $nC0:$

$$2C0 = \frac{2!}{0!(2-0)!} = 1$$

$$\Rightarrow 1 \cdot x^2 = x^2$$

For $nC1:$

$$2C1 = \frac{2!}{1!(2-1)!} = 2$$

$$\Rightarrow 2 \cdot x \cdot y = 2xy$$

For $nC2:$

$$2C2 = \frac{2!}{2!(2-2)!} = 1$$

$$\Rightarrow 1 \cdot y^2 = y^2$$

Therefore, the expansion is: $x^2 – 2xy + y^2$

Key Concept: Complex Expansion, Advanced Coefficient Problem

b) $54a^4 – 144a^3$

[Solution Description] To find the coefficient of $x^4$ in the expansion of $(3x^2 – 2ax + a^2)^4$, we need to consider terms that contribute to $x^4$. The power of each term must sum to 4, given by:

$$(3x^2)^{k_1}(-2ax)^{k_2}(a^2)^{k_3}, \text{ where } k_1 + k_2 + k_3 = 4.$$

We want $2k_1 + k_2 = 4$ for the power of $x^4$.

Case 1: $k_1 = 2$, $k_2 = 0$, $k_3 = 2$

$$\text{Coefficient} = \binom{4}{2,0,2} \cdot (3x^2)^2 \cdot (a^2)^2 = 6 \times 9 \times a^4 = 54a^4$$

Case 2: $k_1 = 1$, $k_2 = 2$, $k_3 = 1$

$$\text{Coefficient} = \binom{4}{1,2,1} \cdot (3x^2)(-2ax)^2(a^2) = 12 \times 3 \times 4a^2x^2 \times a^2 = -144a^3$$

Therefore, the combined coefficient of $x^4$ is $54a^4 – 144a^3$.

Key Concept: Complex Binomial Expansion, Multi-step Solutions

c) 1

[Solution Description] To find the coefficient of the term independent of $x$ in the expansion of $(2x-y)^8$, identify the term where $x$ has a power of zero.

Consider general term form:

$$T_k = \binom{8}{k} (2x)^{8-k} (-y)^k$$

We need power of $x$ to be zero:

$$8-k = 0 \implies k = 8$$

Substitute $k = 8$ in $T_k$:

$$T_8 = \binom{8}{8} (2x)^0 (-y)^8 = 1 \times 1 \times (-y)^8 = y^8$$

So the coefficient of $y^8$ is: $(-1)^8 = 1$

Key Concept: Basic Expansion

a) $x^3 + 3x^2 + 3x + 1$

[Solution Description] Using the binomial theorem, $$(x + 1)^3 = 3C0 x^3 + 3C1 x^2 \cdot 1 + 3C2 x \cdot 1^2 + 3C3 \cdot 1^3$$.

Calculating each term:

The first term: $3C0 = 1$, so we have $$1 \cdot x^3 = x^3$$.

The second term: $3C1 = 3$, so we have $$3 \cdot x^2 \cdot 1 = 3x^2$$.

The third term: $3C2 = 3$, so we have $$3 \cdot x \cdot 1^2 = 3x$$.

The fourth term: $3C3 = 1$, so we have $$1 \cdot 1^3 = 1$$.

Thus, the expansion is $$x^3 + 3x^2 + 3x + 1$$.

Key Concept: Basic Expansion

b) $1 + 3x + 3x^2 + x^3$

[Solution Description] To expand $(1 + x)^3$ using the binomial theorem, we use the formula:

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

Here, $a = 1$, $b = x$, and $n = 3$. So,

$$(1 + x)^3 = \sum_{r=0}^{3} \binom{3}{r} 1^{3-r} x^r$$

Calculating each term:

$$\binom{3}{0} = 1, \quad \text{term is } 1^3x^0 = 1$$

$$\binom{3}{1} = 3, \quad \text{term is } 1^2x^1 = 3x$$

$$\binom{3}{2} = 3, \quad \text{term is } 1^1x^2 = 3x^2$$

$$\binom{3}{3} = 1, \quad \text{term is } 1^0x^3 = x^3$$

Adding these terms: $1 + 3x + 3x^2 + x^3$

Key Concept: Pascal’s Triangle Definition

b) A triangular array of numbers used for binomial coefficients

[Solution Description] Pascal’s triangle is an array of numbers where each number is the sum of the two numbers directly above it, used to determine coefficients in the expansion of a binomial expression.

Key Concept: Negative Binomial Expansion

a) $16x^4 – 32x^3y + 24x^2y^2 – 8xy^3 + y^4$

[Solution Description] The binomial theorem gives us $$(a – b)^n = \sum_{k=0}^{n} {^nC_k a^{n-k}(-b)^k}$$. Here $a = 2x$, $b = y$, and $n = 4$. Thus we have:

$${(2x – y)^4} = {^4C_0 (2x)^4 (-y)^0} + {^4C_1 (2x)^3 (-y)^1} + {^4C_2 (2x)^2 (-y)^2}$$

$$+ {^4C_3 (2x)^1 (-y)^3} + {^4C_4 (2x)^0 (-y)^4}$$

$$= 1 \cdot 16x^4 – 4 \cdot 8x^3 \cdot y + 6 \cdot 4x^2 \cdot y^2 – 4 \cdot 2x \cdot y^3 + 1 \cdot y^4$$

$$= 16x^4 – 32x^3y + 24x^2y^2 – 8xy^3 + y^4$$

Therefore, the expansion is $$16x^4 – 32x^3y + 24x^2y^2 – 8xy^3 + y^4$$.

Key Concept: Expansion Application

d) $1080a^3b^2$

[Solution Description] In the expansion of $(3a – 2b)^5$, the general term is given by:

$$T_k = \binom{5}{k} (3a)^{5-k} (-2b)^{k}$$

We need to find the term where the powers match $a^3b^2$. Equating the powers, we get:

$$5-k = 3$$

$$k = 2$$

Substituting $k = 2$ into the expression for $T_k$, we have:

$$T_2 = \binom{5}{2} (3a)^3 (-2b)^2 = 10 \times 27a^3 \times 4b^2 = 1080a^3b^2$$

Therefore, the required term is $1080a^3b^2$.

Key Concept: Identify Terms

c) 600

[Solution Description] The general term in the expansion is given by:

$$T_k = \binom{4}{k} (2x)^{4-k} (-5)^k$$

We need to find the term containing $x^2$, which means $4-k=2$, so $k=2$.

Now calculate:

$$T_2 = \binom{4}{2} (2x)^{2} (-5)^{2} = 6 \cdot 4x^2 \cdot 25 = 600x^2$$

Therefore, the coefficient of $x^2$ is 600.

Key Concept: Term Count

b) 6

[Solution Description] The number of terms in the expansion of a binomial expression $(a + b)^n$ is $n + 1$. For $(p + q)^5$, this gives:

$5 + 1 = 6$

Therefore, there are 6 terms in the expansion.

Key Concept: Basic Expansion

a) $1 + 3x + 3x^2 + x^3$

[Solution Description] Use the binomial theorem: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$. Here $a=1, b=x, n=3$.

Calculate each term:

$$\text{First term: } \binom{3}{0}(1)^3(x)^0 = 1$$

$$\text{Second term: } \binom{3}{1}(1)^2(x)^1 = 3x$$

$$\text{Third term: } \binom{3}{2}(1)^1(x)^2 = 3x^2$$

$$\text{Fourth term: } \binom{3}{3}(1)^0(x)^3 = x^3$$

Combine these to get: $$1 + 3x + 3x^2 + x^3$$.

Key Concept: Binomial Coefficients

a) 4C2

[Solution Description] In a binomial expansion, the term containing $x^k$ has the binomial coefficient $$\binom{n}{k}$$. Here, $n=4$ and $k=2$, so the binomial coefficient is $$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$$.

Key Concept: Basic Expansion

a) $a^3 + 3a^2b + 3ab^2 + b^3$

[Solution Description] Using the binomial theorem, the expansion of $(a + b)^3$ is given by:

$$(a + b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} b^k$$

Calculating each term:

– For $k=0$: $$\binom{3}{0} a^3 b^0 = 1 \cdot a^3 = a^3$$

– For $k=1$: $$\binom{3}{1} a^2 b^1 = 3 \cdot a^2 b$$

– For $k=2$: $$\binom{3}{2} a^1 b^2 = 3 \cdot ab^2$$

– For $k=3$: $$\binom{3}{3} a^0 b^3 = 1 \cdot b^3 = b^3$$

Therefore, the expansion is $$a^3 + 3a^2b + 3ab^2 + b^3$$.

Key Concept: Binomial Coefficients

c) 35

[Solution Description] In the binomial expansion of $(a+b)^7$, each term can be represented as $$\binom{7}{k} a^{7-k} b^k$$. For the term $a^4b^3$, we have $7-k = 4$ and $k = 3$. Hence, the binomial coefficient is $\binom{7}{3}$.

$$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$

Therefore, the binomial coefficient is 35.

Key Concept: Pascal’s Triangle

b) 1, 5, 10, 10, 5, 1

[Solution Description] The coefficients for the expansion of $(a + b)^5$ are found in the sixth row of Pascal’s Triangle (starting with the top row as row zero). These coefficients are:

$$1, 5, 10, 10, 5, 1$$

Each entry corresponds to a binomial coefficient $^5C_r$, where $r$ goes from 0 to 5. Therefore, the row corresponding to the expansion of $(a + b)^5$ is 1, 5, 10, 10, 5, 1.

Key Concept: Basic Expansion

a) $1 – 3x + 3x^2 – x^3$

[Solution Description] According to the binomial theorem, the expansion of $(a – b)^n$ is given by:

$$(a – b)^n = nC0 \cdot a^n – nC1 \cdot a^{n-1} \cdot b + nC2 \cdot a^{n-2} \cdot b^2 – … + (-1)^n \cdot nCn \cdot b^n$$

For $a = 1$, $b = x$, and $n = 3$, the expansion becomes:

$$(1 – x)^3 = 3C0 \cdot 1^3 – 3C1 \cdot 1^2 \cdot x + 3C2 \cdot 1^1 \cdot x^2 – 3C3 \cdot 1^0 \cdot x^3$$

Calculating each term:

The first term is $$3C0 \cdot 1^3 = 1$$.

The second term is $$-3C1 \cdot 1^2 \cdot x = -3x$$.

The third term is $$3C2 \cdot 1^1 \cdot x^2 = 3x^2$$.

The fourth term is $$-3C3 \cdot 1^0 \cdot x^3 = -x^3$$.

Thus, the expansion is $$1 – 3x + 3x^2 – x^3$$.

Key Concept: Basic Expansion

b) $x^2 – 2x + 1$

[Solution Description] Using the binomial theorem, the expansion of $(x – 1)^2$ is given by:

$$(x-1)^2 = x^2 – 2x \cdot 1 + 1^2 = x^2 – 2x + 1$$

Key Concept: Coefficient Calculation

b) 40

[Solution Description] The binomial expansion of $(2 + x)^5$ can be written using the formula:

$$(2 + x)^5 = \sum_{r=0}^{5} \binom{5}{r} (2)^{5-r} x^r$$

We need to find the coefficient of $x^3$. Here, $a = 2$, $b = x$, $n = 5$, and we are looking for the term where $r = 3$.

The coefficient for $x^3$ will be:

$$\binom{5}{3} \cdot 2^{5-3} = \frac{5!}{3!(5-3)!} \cdot 2^2$$

Calculating further:

$$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$

And,

$$2^2 = 4$$

Thus, the coefficient of $x^3$ is:

$$10 \times 4 = 40$$

Key Concept: Pascal’s Triangle

b) 4

[Solution Description] The coefficients for $(a + b)^4$ are 1, 4, 6, 4, 1 from Pascal’s Triangle. The coefficient of $a^3b$ corresponds to the second term which is 4.

Key Concept: Basic Expansion

b) $1 + 3x + 3x^2 + x^3$

[Solution Description] Using the binomial theorem, the expansion of $(a + b)^n$ is given by:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Here, $a = 1$, $b = x$, and $n = 3$. Thus,

$$(1 + x)^3 = \binom{3}{0} \cdot 1^{3-0} \cdot x^0 + \binom{3}{1} \cdot 1^{3-1} \cdot x^1$$

$$\quad + \binom{3}{2} \cdot 1^{3-2} \cdot x^2 + \binom{3}{3} \cdot 1^{3-3} \cdot x^3$$

$$= 1 + 3x + 3x^2 + x^3$$

Therefore, the expansion is $$1 + 3x + 3x^2 + x^3$$.

Key Concept: Term Count

b) 5

[Solution Description] The number of terms in the expansion of a binomial expression $(x+y)^n$ is given by $n+1$. Here, $n=4$, so we have:

$$n + 1 = 4 + 1 = 5$$

Therefore, there are 5 terms in the expansion of $(x + y)^4$.

Key Concept: Term Count

c) 7

[Solution Description] According to the properties of binomial expansion, the number of terms is $$n+1$$. For $$(a + b)^6$$, this results in $$6 + 1 = 7$$ terms.