Binomial TheoremClass 11 Mathematics NCERT Solutions

Given: The expression $(1-2x)^5$.

To Find: The expansion of the given expression.

Formula: The binomial theorem for $(a-b)^n$ is: $(a-b)^n = \sum_{r=0}^{n} (-1)^r \,^nC_r a^{n-r} b^r$ $(a-b)^n = \,^nC_0 a^n - \,^nC_1 a^{n-1}b + \,^nC_2 a^{n-2}b^2 - \dots + (-1)^n \,^nC_n b^n$

Solution: Here, $a=1$, $b=2x$, and $n=5$.

We know that: $\,^5C_0 = 1$ $\,^5C_1 = 5$ $\,^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$ $\,^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$ $\,^5C_4 = 5$ $\,^5C_5 = 1$

Substituting these values: $(1-2x)^5 = 1(1) - 5(1)(2x) + 10(1)(4x^2) - 10(1)(8x^3) + 5(1)(16x^4) - 1(32x^5)$ $= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$

Final Answer: The expansion of $(1-2x)^5$ is $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$.

Given: The expression $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$.

To Find: The expansion of the given expression.

Formula: The binomial theorem for $(a-b)^n$ is: $(a-b)^n = \,^nC_0 a^n - \,^nC_1 a^{n-1}b + \,^nC_2 a^{n-2}b^2 - \dots + (-1)^n \,^nC_n b^n$

Solution: Here, $a=\frac{2}{x}$, $b=\frac{x}{2}$, and $n=5$.

Using the coefficients $\,^5C_0=1, \,^5C_1=5, \,^5C_2=10, \,^5C_3=10, \,^5C_4=5, \,^5C_5=1$: $= 1\left(\frac{32}{x^5}\right) - 5\left(\frac{16}{x^4}\right)\left(\frac{x}{2}\right) + 10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right) - 10\left(\frac{4}{x^2}\right)\left(\frac{x^3}{8}\right) + 5\left(\frac{2}{x}\right)\left(\frac{x^4}{16}\right) - 1\left(\frac{x^5}{32}\right)$ $= \frac{32}{x^5} - 5\left(\frac{8}{x^3}\right) + 10\left(\frac{2}{x}\right) - 10\left(\frac{x}{2}\right) + 5\left(\frac{x^3}{8}\right) - \frac{x^5}{32}$ $= \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5}{8}x^3 - \frac{x^5}{32}$

Final Answer: The expansion of $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$ is $\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5}{8}x^3 - \frac{x^5}{32}$.

Given: The expression $(2x-3)^6$.

To Find: The expansion of the given expression.

Formula: The binomial theorem for $(a-b)^n$ is: $(a-b)^n = \sum_{r=0}^{n} (-1)^r \,^nC_r a^{n-r} b^r$

Solution: Here, $a=2x$, $b=3$, and $n=6$.

We know that: $\,^6C_0 = 1, \,^6C_1 = 6, \,^6C_2 = 15, \,^6C_3 = 20, \,^6C_4 = 15, \,^6C_5 = 6, \,^6C_6 = 1$

Substituting these values: $= 1(64x^6) - 6(32x^5)(3) + 15(16x^4)(9) - 20(8x^3)(27) + 15(4x^2)(81) - 6(2x)(243) + 1(729)$ $= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 6075x^2 - 2916x + 729$

Final Answer: The expansion of $(2x-3)^6$ is $64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 6075x^2 - 2916x + 729$.

Given: The expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$.

To Find: The expansion of the given expression.

Formula: The binomial theorem for $(a+b)^n$ is: $(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$

Solution: Here, $a=\frac{x}{3}$, $b=\frac{1}{x}$, and $n=5$.

Using the coefficients $\,^5C_0=1, \,^5C_1=5, \,^5C_2=10, \,^5C_3=10, \,^5C_4=5, \,^5C_5=1$: $= 1\left(\frac{x^5}{243}\right) + 5\left(\frac{x^4}{81}\right)\left(\frac{1}{x}\right) + 10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right) + 10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right) + 5\left(\frac{x}{3}\right)\left(\frac{1}{x^4}\right) + 1\left(\frac{1}{x^5}\right)$ $= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$

Final Answer: The expansion of $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$ is $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$.

Given: The expression $\left(x+\frac{1}{x}\right)^{6}$.

To Find: The expansion of the given expression.

Formula: The binomial theorem for $(a+b)^n$ is: $(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$

Solution: Here, $a=x$, $b=\frac{1}{x}$, and $n=6$.

Using the coefficients $\,^6C_0=1, \,^6C_1=6, \,^6C_2=15, \,^6C_3=20, \,^6C_4=15, \,^6C_5=6, \,^6C_6=1$: $= 1(x^6) + 6(x^5)\left(\frac{1}{x}\right) + 15(x^4)\left(\frac{1}{x^2}\right) + 20(x^3)\left(\frac{1}{x^3}\right) + 15(x^2)\left(\frac{1}{x^4}\right) + 6(x)\left(\frac{1}{x^5}\right) + 1\left(\frac{1}{x^6}\right)$ $= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$

Final Answer: The expansion of $\left(x+\frac{1}{x}\right)^{6}$ is $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$.

Given: The expression $(96)^3$.

To Find: The value of the expression using the binomial theorem.

Solution: First, we express 96 as a difference of two numbers whose powers are easy to calculate. We can write $96 = 100 - 4$. So, $(96)^3 = (100-4)^3$.

Using the binomial expansion for $(a-b)^3 = \,^3C_0 a^3 - \,^3C_1 a^2b + \,^3C_2 ab^2 - \,^3C_3 b^3$: Here, $a=100$ and $b=4$.

$(100-4)^3 = \,^3C_0 (100)^3 - \,^3C_1 (100)^2(4) + \,^3C_2 (100)(4)^2 - \,^3C_3 (4)^3$ $= 1(1000000) - 3(10000)(4) + 3(100)(16) - 1(64)$ $= 1000000 - 120000 + 4800 - 64$ $= (1000000 + 4800) - (120000 + 64)$ $= 1004800 - 120064$ $= 884736$

Final Answer: The value of $(96)^3$ is $884,736$.

Given: The expression $(102)^5$.

To Find: The value of the expression using the binomial theorem.

Solution: First, we express 102 as a sum of two numbers. We can write $102 = 100 + 2$. So, $(102)^5 = (100+2)^5$.

Using the binomial expansion for $(a+b)^5$: $(a+b)^5 = \,^5C_0 a^5 + \,^5C_1 a^4b + \,^5C_2 a^3b^2 + \,^5C_3 a^2b^3 + \,^5C_4 ab^4 + \,^5C_5 b^5$ Here, $a=100$ and $b=2$.

$(100+2)^5 = \,^5C_0(100)^5 + \,^5C_1(100)^4(2) + \,^5C_2(100)^3(2)^2 + \,^5C_3(100)^2(2)^3 + \,^5C_4(100)(2)^4 + \,^5C_5(2)^5$ $= 1(100)^5 + 5(100)^4(2) + 10(100)^3(4) + 10(100)^2(8) + 5(100)(16) + 1(32)$ $= 1(10,000,000,000) + 10(100,000,000) + 40(1,000,000) + 80(10,000) + 8000 + 32$ $= 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32$ $= 11,040,808,032$

Final Answer: The value of $(102)^5$ is $11,040,808,032$.

Given: The expression $(101)^4$.

To Find: The value of the expression using the binomial theorem.

Solution: First, we express 101 as a sum. We can write $101 = 100 + 1$. So, $(101)^4 = (100+1)^4$.

Using the binomial expansion for $(a+b)^4$: $(a+b)^4 = \,^4C_0 a^4 + \,^4C_1 a^3b + \,^4C_2 a^2b^2 + \,^4C_3 ab^3 + \,^4C_4 b^4$ Here, $a=100$ and $b=1$.

$(100+1)^4 = \,^4C_0(100)^4 + \,^4C_1(100)^3(1) + \,^4C_2(100)^2(1)^2 + \,^4C_3(100)(1)^3 + \,^4C_4(1)^4$ $= 1(100)^4 + 4(100)^3 + 6(100)^2 + 4(100) + 1$ $= 1(100,000,000) + 4(1,000,000) + 6(10,000) + 400 + 1$ $= 100,000,000 + 4,000,000 + 60,000 + 400 + 1$ $= 104,060,401$

Final Answer: The value of $(101)^4$ is $104,060,401$.

Given: The expression $(99)^5$.

To Find: The value of the expression using the binomial theorem.

Solution: First, we express 99 as a difference. We can write $99 = 100 - 1$. So, $(99)^5 = (100-1)^5$.

Using the binomial expansion for $(a-b)^5$: $(a-b)^5 = \,^5C_0 a^5 - \,^5C_1 a^4b + \,^5C_2 a^3b^2 - \,^5C_3 a^2b^3 + \,^5C_4 ab^4 - \,^5C_5 b^5$ Here, $a=100$ and $b=1$.

$(100-1)^5 = \,^5C_0(100)^5 - \,^5C_1(100)^4(1) + \,^5C_2(100)^3(1)^2 - \,^5C_3(100)^2(1)^3 + \,^5C_4(100)(1)^4 - \,^5C_5(1)^5$ $= 1(100)^5 - 5(100)^4 + 10(100)^3 - 10(100)^2 + 5(100) - 1$ $= 1(10,000,000,000) - 5(100,000,000) + 10(1,000,000) - 10(10,000) + 500 - 1$ $= 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$

Group positive and negative terms: Positive sum = $10,000,000,000 + 10,000,000 + 500 = 10,010,000,500$ Negative sum = $-500,000,000 - 100,000 - 1 = -500,100,001$

Total = $10,010,000,500 - 500,100,001 = 9,509,900,499$

Final Answer: The value of $(99)^5$ is $9,509,900,499$.

Given: Two numbers, $(1.1)^{10000}$ and $1000$.

To Find: Which of the two numbers is larger.

Solution: We can write $1.1$ as $(1 + 0.1)$. So, $(1.1)^{10000} = (1 + 0.1)^{10000}$.

Using the binomial theorem for $(1+x)^n = \,^nC_0 + \,^nC_1 x + \,^nC_2 x^2 + \dots + \,^nC_n x^n$. Here, $x=0.1$ and $n=10000$.

Let's evaluate the first two terms: $\,^{10000}C_0 = 1$ $\,^{10000}C_1(0.1) = 10000 \times 0.1 = 1000$

So, the expansion is: $(1.1)^{10000} = 1 + 1000 + \,^{10000}C_2(0.1)^2 + \dots$ $= 1001 + (\text{sum of other positive terms})$

Since all the remaining terms in the expansion ($\,^{10000}C_2(0.1)^2$, etc.) are positive, the sum will be greater than 1001. Therefore, $(1.1)^{10000} > 1001$. Since $1001 > 1000$, it follows that $(1.1)^{10000} > 1000$.

Final Answer: $(1.1)^{10000}$ is larger than $1000$.

Solution: First, we expand $(a+b)^4$ and $(a-b)^4$ using the binomial theorem.

$(a+b)^4 = \,^4C_0 a^4 + \,^4C_1 a^3b + \,^4C_2 a^2b^2 + \,^4C_3 ab^3 + \,^4C_4 b^4$ $= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

$(a-b)^4 = \,^4C_0 a^4 - \,^4C_1 a^3b + \,^4C_2 a^2b^2 - \,^4C_3 ab^3 + \,^4C_4 b^4$ $= a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$

Now, we subtract the second expansion from the first: $(a+b)^4 - (a-b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$ $= (a^4 - a^4) + (4a^3b - (-4a^3b)) + (6a^2b^2 - 6a^2b^2) + (4ab^3 - (-4ab^3)) + (b^4 - b^4)$ $= 8a^3b + 8ab^3$ $= 8ab(a^2 + b^2)$

Solution: We use the result from Part 1 by substituting $a = \sqrt{3}$ and $b = \sqrt{2}$.

$a = \sqrt{3} \implies a^2 = 3$ $b = \sqrt{2} \implies b^2 = 2$

Using the formula $8ab(a^2+b^2)$: $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4} = 8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2)$ $= 8\sqrt{6}(3+2)$ $= 8\sqrt{6}(5)$ $= 40\sqrt{6}$

Final Answer: $(a+b)^{4}-(a-b)^{4} = 8ab(a^2 + b^2)$, and the value of $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$ is $40\sqrt{6}$.

Solution: First, we expand $(x+1)^6$ and $(x-1)^6$ using the binomial theorem.

Now, we add the two expansions. The terms with odd powers will cancel out. $(x+1)^6 + (x-1)^6 = 2(\,^6C_0 x^6 + \,^6C_2 x^4 + \,^6C_4 x^2 + \,^6C_6)$

We calculate the coefficients: $\,^6C_0 = 1$ $\,^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$ $\,^6C_4 = \,^6C_{6-4} = \,^6C_2 = 15$ $\,^6C_6 = 1$

Substituting these values: $(x+1)^6 + (x-1)^6 = 2(1 \cdot x^6 + 15 \cdot x^4 + 15 \cdot x^2 + 1)$ $= 2(x^6 + 15x^4 + 15x^2 + 1)$

Solution: This expression matches the form from Part 1 with $x=\sqrt{2}$. We substitute $x=\sqrt{2}$ into the simplified expression $2(x^6 + 15x^4 + 15x^2 + 1)$.

If $x = \sqrt{2}$: $x^2 = (\sqrt{2})^2 = 2$ $x^4 = (x^2)^2 = 2^2 = 4$ $x^6 = (x^2)^3 = 2^3 = 8$

Substitute these into the expression: $2(8 + 15(4) + 15(2) + 1)$ $= 2(8 + 60 + 30 + 1)$ $= 2(99)$ $= 198$

Final Answer: $(x+1)^{6}+(x-1)^{6} = 2(x^6 + 15x^4 + 15x^2 + 1)$, and the value of $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$ is $198$.

To Prove: $9^{n+1}-8n-9$ is divisible by 64 for any positive integer $n$.

Proof: We can write $9$ as $(1+8)$. Therefore, $9^{n+1} = (1+8)^{n+1}$.

Using the binomial theorem to expand $(1+8)^{n+1}$: $(1+8)^{n+1} = \,^{n+1}C_0 (1)^{n+1} + \,^{n+1}C_1 (1)^n (8)^1 + \,^{n+1}C_2 (1)^{n-1} (8)^2 + \,^{n+1}C_3 (1)^{n-2} (8)^3 + \dots + \,^{n+1}C_{n+1} (8)^{n+1}$

Let's evaluate the first few terms: $\,^{n+1}C_0 = 1$ $\,^{n+1}C_1 = n+1$

So, the expansion becomes: $9^{n+1} = 1 + (n+1)(8) + \,^{n+1}C_2 (8^2) + \,^{n+1}C_3 (8^3) + \dots + (8)^{n+1}$ $9^{n+1} = 1 + 8n + 8 + \,^{n+1}C_2 (64) + \,^{n+1}C_3 (512) + \dots$ $9^{n+1} = 8n + 9 + 64 \,^{n+1}C_2 + 64(8) \,^{n+1}C_3 + \dots$

We can factor out 64 from the third term onwards: $9^{n+1} = 8n + 9 + 64 \left[ \,^{n+1}C_2 + 8 \,^{n+1}C_3 + \dots + 8^{n-1} \,^{n+1}C_{n+1} \right]$

Rearranging the terms: $9^{n+1} - 8n - 9 = 64 \left[ \,^{n+1}C_2 + 8 \,^{n+1}C_3 + \dots \right]$

Let $k = \,^{n+1}C_2 + 8 \,^{n+1}C_3 + \dots$. Since $n$ is a positive integer, all the combination terms are integers, so $k$ is an integer. Therefore, $9^{n+1} - 8n - 9 = 64k$, where $k$ is an integer.

This shows that $9^{n+1} - 8n - 9$ is a multiple of 64.

To Prove: $\sum_{r=0}^{n} 3^{r} \,^nC_r = 4^n$.

Proof: We start with the general formula for the binomial theorem: $(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$

Let's compare the given summation with the general formula. The given summation is $\sum_{r=0}^{n} \,^nC_r 3^r$. We can write this as $\sum_{r=0}^{n} \,^nC_r (1)^{n-r} (3)^r$, since $1^{n-r} = 1$ for any value of $n-r$.

Now, the expression $\sum_{r=0}^{n} \,^nC_r (1)^{n-r} (3)^r$ perfectly matches the expansion of $(a+b)^n$ with $a=1$ and $b=3$.

Therefore, $\sum_{r=0}^{n} 3^{r} \,^nC_r = (1+3)^n$ $= 4^n$

Thus, LHS = $\sum_{r=0}^{n} 3^{r} \,^nC_r = 4^n$ = RHS.

To Prove: $a-b$ is a factor of $a^n - b^n$ for any positive integer $n$.

Proof: Following the hint, we write $a$ as $a = (a-b) + b$. Then, $a^n = ((a-b) + b)^n$.

We expand this using the binomial theorem, treating $(a-b)$ as the first term and $b$ as the second term. $a^n = \,^nC_0 (a-b)^n b^0 + \,^nC_1 (a-b)^{n-1} b^1 + \,^nC_2 (a-b)^{n-2} b^2 + \dots + \,^nC_{n-1} (a-b)^1 b^{n-1} + \,^nC_n (a-b)^0 b^n$