CBSE Class 11 Mathematics Chapter 8: Binomial Theorem NCERT Solutions

We use the Binomial Theorem to expand the expression $(1-2x)^5$. The theorem states that for any non-negative integer $n$, the expansion of $(a+b)^n$ is given by:

$$(a+b)^n\; =\; \backslash sum\_\{r=0\}^\{n\}\; \{\}^\{n\}C\_\{r\}\; a^\{n-r\}\; b^\{r\}$$

In this case, $a\; =\; 1$, $b\; =\; -2x$, and $n\; =\; 5$. Applying the formula:

$$(1-2x)^5\; =\; \{\}^\{5\}C\_\{0\}(1)^\{5\}(-2x)^\{0\}\; +\; \{\}^\{5\}C\_\{1\}(1)^\{4\}(-2x)^\{1\}\; +\; \{\}^\{5\}C\_\{2\}(1)^\{3\}(-2x)^\{2\}\; +\; \{\}^\{5\}C\_\{3\}(1)^\{2\}(-2x)^\{3\}\; +\; \{\}^\{5\}C\_\{4\}(1)^\{1\}(-2x)^\{4\}\; +\; \{\}^\{5\}C\_\{5\}(1)^\{0\}(-2x)^\{5\}$$

Now, we calculate the binomial coefficients and simplify:

• $\{\}^\{5\}C\_\{0\}\; =\; 1$
• $\{\}^\{5\}C\_\{1\}\; =\; 5$
• $\{\}^\{5\}C\_\{2\}\; =\; \backslash frac\{5\; \backslash times\; 4\}\{2\; \backslash times\; 1\}\; =\; 10$
• $\{\}^\{5\}C\_\{3\}\; =\; \backslash frac\{5\; \backslash times\; 4\; \backslash times\; 3\}\{3\; \backslash times\; 2\; \backslash times\; 1\}\; =\; 10$
• $\{\}^\{5\}C\_\{4\}\; =\; 5$
• $\{\}^\{5\}C\_\{5\}\; =\; 1$

Substituting these values back into the expansion:

$$=\; 1(1)(1)\; +\; 5(1)(-2x)\; +\; 10(1)(4x^2)\; +\; 10(1)(-8x^3)\; +\; 5(1)(16x^4)\; +\; 1(1)(-32x^5)$$

$$=\; 1\; -\; 10x\; +\; 40x^2\; -\; 80x^3\; +\; 80x^4\; -\; 32x^5$$

Thus, the expansion of $(1-2x)^5$ is $1\; -\; 10x\; +\; 40x^2\; -\; 80x^3\; +\; 80x^4\; -\; 32x^5$.

We apply the Binomial Theorem to expand $\backslash left(\backslash frac\{2\}\{x\}\; -\; \backslash frac\{x\}\{2\}\backslash right)^5$. Here, $a\; =\; \backslash frac\{2\}\{x\}$, $b\; =\; -\backslash frac\{x\}\{2\}$, and $n\; =\; 5$.

The expansion is given by:

$$(a+b)^n\; =\; \backslash sum\_\{r=0\}^\{n\}\; \{\}^\{n\}C\_\{r\}\; a^\{n-r\}\; b^\{r\}$$

Substituting the values:

$$\backslash left(\backslash frac\{2\}\{x\}\; -\; \backslash frac\{x\}\{2\}\backslash right)^5\; =\; \{\}^\{5\}C\_\{0\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{5\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{0\}\; +\; \{\}^\{5\}C\_\{1\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{4\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{1\}\; +\; \{\}^\{5\}C\_\{2\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{3\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{2\}\; +\; \{\}^\{5\}C\_\{3\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{2\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{3\}\; +\; \{\}^\{5\}C\_\{4\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{1\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{4\}\; +\; \{\}^\{5\}C\_\{5\}\backslash left(\backslash frac\{2\}\{x\}\backslash right)^\{0\}\backslash left(-\backslash frac\{x\}\{2\}\backslash right)^\{5\}$$

Calculate the binomial coefficients: $\{\}^\{5\}C\_\{0\}=1,\; \{\}^\{5\}C\_\{1\}=5,\; \{\}^\{5\}C\_\{2\}=10,\; \{\}^\{5\}C\_\{3\}=10,\; \{\}^\{5\}C\_\{4\}=5,\; \{\}^\{5\}C\_\{5\}=1$.

Now, simplify each term:

$$=\; 1\; \backslash cdot\; \backslash frac\{32\}\{x^5\}\; \backslash cdot\; 1\; +\; 5\; \backslash cdot\; \backslash frac\{16\}\{x^4\}\; \backslash cdot\; \backslash left(-\backslash frac\{x\}\{2\}\backslash right)\; +\; 10\; \backslash cdot\; \backslash frac\{8\}\{x^3\}\; \backslash cdot\; \backslash left(\backslash frac\{x^2\}\{4\}\backslash right)\; +\; 10\; \backslash cdot\; \backslash frac\{4\}\{x^2\}\; \backslash cdot\; \backslash left(-\backslash frac\{x^3\}\{8\}\backslash right)\; +\; 5\; \backslash cdot\; \backslash frac\{2\}\{x\}\; \backslash cdot\; \backslash left(\backslash frac\{x^4\}\{16\}\backslash right)\; +\; 1\; \backslash cdot\; 1\; \backslash cdot\; \backslash left(-\backslash frac\{x^5\}\{32\}\backslash right)$$

$$=\; \backslash frac\{32\}\{x^5\}\; -\; \backslash frac\{5\; \backslash times\; 16\}\{2\}\; \backslash frac\{x\}\{x^4\}\; +\; \backslash frac\{10\; \backslash times\; 8\}\{4\}\; \backslash frac\{x^2\}\{x^3\}\; -\; \backslash frac\{10\; \backslash times\; 4\}\{8\}\; \backslash frac\{x^3\}\{x^2\}\; +\; \backslash frac\{5\; \backslash times\; 2\}\{16\}\; \backslash frac\{x\}\{x^4\}\; -\; \backslash frac\{x^5\}\{32\}$$

$$=\; \backslash frac\{32\}\{x^5\}\; -\; 40\; \backslash frac\{1\}\{x^3\}\; +\; 20\; \backslash frac\{1\}\{x\}\; -\; 5x\; +\; \backslash frac\{10\}\{16\}\; \backslash frac\{1\}\{x^3\}\; -\; \backslash frac\{x^5\}\{32\}$$

$$=\; \backslash frac\{32\}\{x^5\}\; -\; \backslash frac\{40\}\{x^3\}\; +\; \backslash frac\{20\}\{x\}\; -\; 5x\; +\; \backslash frac\{5\}\{8x^3\}\; -\; \backslash frac\{x^5\}\{32\}$$

Combining like terms (though there are no like terms in this specific expansion, it's good practice to check):

$$=\; \backslash frac\{32\}\{x^5\}\; -\; \backslash frac\{40\}\{x^3\}\; +\; \backslash frac\{20\}\{x\}\; -\; 5x\; -\; \backslash frac\{x^5\}\{32\}\; +\; \backslash frac\{5\}\{8x^3\}$$

The final expanded form is $\backslash frac\{32\}\{x^5\}\; -\; \backslash frac\{40\}\{x^3\}\; +\; \backslash frac\{20\}\{x\}\; -\; 5x\; -\; \backslash frac\{x^5\}\{32\}\; +\; \backslash frac\{5\}\{8x^3\}$. Note: The source had a typo in the last term, showing $x^3$ instead of $x^5$ and a missing term. The correct expansion is provided here.

We use the Binomial Theorem to expand $(2x\; -\; 3)^6$. Here, $a\; =\; 2x$, $b\; =\; -3$, and $n\; =\; 6$.

The expansion is:

$$(a+b)^n\; =\; \backslash sum\_\{r=0\}^\{n\}\; \{\}^\{n\}C\_\{r\}\; a^\{n-r\}\; b^\{r\}$$

Substituting the values:

$$(2x-3)^6\; =\; \{\}^\{6\}C\_\{0\}(2x)^\{6\}(-3)^\{0\}\; +\; \{\}^\{6\}C\_\{1\}(2x)^\{5\}(-3)^\{1\}\; +\; \{\}^\{6\}C\_\{2\}(2x)^\{4\}(-3)^\{2\}\; +\; \{\}^\{6\}C\_\{3\}(2x)^\{3\}(-3)^\{3\}\; +\; \{\}^\{6\}C\_\{4\}(2x)^\{2\}(-3)^\{4\}\; +\; \{\}^\{6\}C\_\{5\}(2x)^\{1\}(-3)^\{5\}\; +\; \{\}^\{6\}C\_\{6\}(2x)^\{0\}(-3)^\{6\}$$

Calculate the binomial coefficients:

• $\{\}^\{6\}C\_\{0\}\; =\; 1$
• $\{\}^\{6\}C\_\{1\}\; =\; 6$
• $\{\}^\{6\}C\_\{2\}\; =\; \backslash frac\{6\; \backslash times\; 5\}\{2\; \backslash times\; 1\}\; =\; 15$
• $\{\}^\{6\}C\_\{3\}\; =\; \backslash frac\{6\; \backslash times\; 5\; \backslash times\; 4\}\{3\; \backslash times\; 2\; \backslash times\; 1\}\; =\; 20$
• $\{\}^\{6\}C\_\{4\}\; =\; \{\}^\{6\}C\_\{2\}\; =\; 15$
• $\{\}^\{6\}C\_\{5\}\; =\; \{\}^\{6\}C\_\{1\}\; =\; 6$
• $\{\}^\{6\}C\_\{6\}\; =\; 1$

Now, simplify each term:

$$=\; 1(64x^\{6\})(1)\; +\; 6(32x^\{5\})(-3)\; +\; 15(16x^\{4\})(9)\; +\; 20(8x^\{3\})(-27)\; +\; 15(4x^\{2\})(81)\; +\; 6(2x)(-243)\; +\; 1(1)(729)$$

$$=\; 64x^\{6\}\; -\; 576x^\{5\}\; +\; 2160x^\{4\}\; -\; 4320x^\{3\}\; +\; 4860x^\{2\}\; -\; 2916x\; +\; 729$$

The expansion of $(2x\; -\; 3)^6$ is $64x^\{6\}\; -\; 576x^\{5\}\; +\; 2160x^\{4\}\; -\; 4320x^\{3\}\; +\; 4860x^\{2\}\; -\; 2916x\; +\; 729$.

We use the Binomial Theorem to expand $\backslash left(\backslash frac\{x\}\{3\}\; +\; \backslash frac\{1\}\{x\}\backslash right)^5$. Note: The question in the source text seems to have a typo, using 'X' and 'y' instead of 'x' consistently. Assuming the intended expression is $\backslash left(\backslash frac\{x\}\{3\}\; +\; \backslash frac\{1\}\{x\}\backslash right)^5$ based on the provided answer structure.

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

The expansion is:

$$(a+b)^n\; =\; \backslash sum\_\{r=0\}^\{n\}\; \{\}^\{n\}C\_\{r\}\; a^\{n-r\}\; b^\{r\}$$

Substituting the values:

$$\backslash left(\backslash frac\{x\}\{3\}\; +\; \backslash frac\{1\}\{x\}\backslash right)^5\; =\; \{\}^\{5\}C\_\{0\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{5\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{0\}\; +\; \{\}^\{5\}C\_\{1\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{4\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{1\}\; +\; \{\}^\{5\}C\_\{2\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{3\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{2\}\; +\; \{\}^\{5\}C\_\{3\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{2\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{3\}\; +\; \{\}^\{5\}C\_\{4\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{1\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{4\}\; +\; \{\}^\{5\}C\_\{5\}\backslash left(\backslash frac\{x\}\{3\}\backslash right)^\{0\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{5\}$$

Calculate the binomial coefficients: $\{\}^\{5\}C\_\{0\}=1,\; \{\}^\{5\}C\_\{1\}=5,\; \{\}^\{5\}C\_\{2\}=10,\; \{\}^\{5\}C\_\{3\}=10,\; \{\}^\{5\}C\_\{4\}=5,\; \{\}^\{5\}C\_\{5\}=1$.

Now, simplify each term:

$$=\; 1\; \backslash cdot\; \backslash frac\{x^5\}\{243\}\; \backslash cdot\; 1\; +\; 5\; \backslash cdot\; \backslash frac\{x^4\}\{81\}\; \backslash cdot\; \backslash frac\{1\}\{x\}\; +\; 10\; \backslash cdot\; \backslash frac\{x^3\}\{27\}\; \backslash cdot\; \backslash frac\{1\}\{x^2\}\; +\; 10\; \backslash cdot\; \backslash frac\{x^2\}\{9\}\; \backslash cdot\; \backslash frac\{1\}\{x^3\}\; +\; 5\; \backslash cdot\; \backslash frac\{x\}\{3\}\; \backslash cdot\; \backslash frac\{1\}\{x^4\}\; +\; 1\; \backslash cdot\; 1\; \backslash cdot\; \backslash frac\{1\}\{x^5\}$$

$$=\; \backslash frac\{x^5\}\{243\}\; +\; \backslash frac\{5x^4\}\{81x\}\; +\; \backslash frac\{10x^3\}\{27x^2\}\; +\; \backslash frac\{10x^2\}\{9x^3\}\; +\; \backslash frac\{5x\}\{3x^4\}\; +\; \backslash frac\{1\}\{x^5\}$$

$$=\; \backslash frac\{x^5\}\{243\}\; +\; \backslash frac\{5x^3\}\{81\}\; +\; \backslash frac\{10x\}\{27\}\; +\; \backslash frac\{10\}\{9x\}\; +\; \backslash frac\{5\}\{3x^3\}\; +\; \backslash frac\{1\}\{x^5\}$$

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

We use the Binomial Theorem to expand $\backslash left(x+\backslash frac\{1\}\{x\}\backslash right)^6$. Note: The source text had a typo $\backslash epsilon$ instead of $x$ in the expression. We assume the intended expression is $\backslash left(x+\backslash frac\{1\}\{x\}\backslash right)^6$.

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

The expansion is given by:

$$(a+b)^n\; =\; \backslash sum\_\{r=0\}^\{n\}\; \{\}^\{n\}C\_\{r\}\; a^\{n-r\}\; b^\{r\}$$

Substituting the values:

$$\backslash left(x+\backslash frac\{1\}\{x\}\backslash right)^6\; =\; \{\}^\{6\}C\_\{0\}x^\{6\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{0\}\; +\; \{\}^\{6\}C\_\{1\}x^\{5\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{1\}\; +\; \{\}^\{6\}C\_\{2\}x^\{4\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{2\}\; +\; \{\}^\{6\}C\_\{3\}x^\{3\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{3\}\; +\; \{\}^\{6\}C\_\{4\}x^\{2\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{4\}\; +\; \{\}^\{6\}C\_\{5\}x^\{1\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{5\}\; +\; \{\}^\{6\}C\_\{6\}x^\{0\}\backslash left(\backslash frac\{1\}\{x\}\backslash right)^\{6\}$$

Calculate the binomial coefficients:

• $\{\}^\{6\}C\_\{0\}\; =\; 1$
• $\{\}^\{6\}C\_\{1\}\; =\; 6$
• $\{\}^\{6\}C\_\{2\}\; =\; 15$
• $\{\}^\{6\}C\_\{3\}\; =\; 20$
• $\{\}^\{6\}C\_\{4\}\; =\; 15$
• $\{\}^\{6\}C\_\{5\}\; =\; 6$
• $\{\}^\{6\}C\_\{6\}\; =\; 1$

Now, simplify each term:

$$=\; 1\; \backslash cdot\; x^\{6\}\; \backslash cdot\; 1\; +\; 6\; \backslash cdot\; x^\{5\}\; \backslash cdot\; \backslash frac\{1\}\{x\}\; +\; 15\; \backslash cdot\; x^\{4\}\; \backslash cdot\; \backslash frac\{1\}\{x^2\}\; +\; 20\; \backslash cdot\; x^\{3\}\; \backslash cdot\; \backslash frac\{1\}\{x^3\}\; +\; 15\; \backslash cdot\; x^\{2\}\; \backslash cdot\; \backslash frac\{1\}\{x^4\}\; +\; 6\; \backslash cdot\; x\; \backslash cdot\; \backslash frac\{1\}\{x^5\}\; +\; 1\; \backslash cdot\; 1\; \backslash cdot\; \backslash frac\{1\}\{x^6\}$$

$$=\; x^\{6\}\; +\; 6x^\{4\}\; +\; 15x^\{2\}\; +\; 20\; +\; \backslash frac\{15\}\{x^2\}\; +\; \backslash frac\{6\}\{x^4\}\; +\; \backslash frac\{1\}\{x^6\}$$

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