mathematics-class-11-unit-09-chapter-02-binomial-theorem-l-4-9-fpmf-zalzjw

### <Success>Some important results</Success>
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<p>$ (x+y)^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1} y+\cdots \cdot{ }^n C_n y^n $</p>
<p><strong>1.</strong>  $ { }^n C_0+{ }^n C_1+{ }^n C_2+\cdots \cdot{ }^n C_n=?(1+1)^n=2^n $</p>
<p>Let $x= 1, \ \ y = 1 $ then,<br>
$ { }^n C_0+{ }^n C_1+{ }^n C_2+\cdots \cdot{ }^n C_n=(1+1)^n=2^n $</p>
<p><strong>2.</strong>  Let $x= 1, \ \ y = -1 $ then,</p>
<p>$ { }^n C_0+{ }^n C_1(-1)+{ }^n C_2(-1)^2+\cdots \cdot{ }^n C_n(-1)^n=(1-1)^n=0 $</p>
<p>$ 0 ={ }^n C_0-{ }^n C_1+{ }^n C_2-{ }^n C_3+{ }^n C_4 \ldots(-1)^n{ }^n C_n $</p>
<p>Now, there are two cases:
$n$ is even and $n$ is odd.</p>
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<h3 id="successresult-2success"><Success>Result 2</Success></h3>
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<p>$\begin{aligned} & (1-1)^{2 n}={ }^{2 n} C_0-{ }^{2 n} C_1+{ }^{2 n} C_2 \cdots \cdot+{ }^{2 n} C_{2 n} \\ & { }^{2 n} C_0+{ }^{2 n} C_2+{ }^{2 n} C_4+\cdots{ }^{2 n} C_{2 n}={ }^{2 n} C_1+{ }^{2 n} C_3+{ }^{2 n} C_5+\cdots{ }^{2 n} C_{2 n} \\ & C_0+{C_1}+{C_2}+\cdots \cdot C_{2 n}=2^{(2 n)} \\ & C_0+C_2+C_4+\ldots C_{2 n}=C_1+C_3+C_5+\cdots C_{2 n-1}=2^{2 n-1}\end{aligned}$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-09-chapter-02-binomial-theorem-L-4-9-fpmf-zalzjw-5.jpg"></p>
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### <Success>Result 3</Success>
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<p>${ }^n C_0^2+{ }^n C_1^2+{ }^n C_2^2+\cdots{ }^n C_n^2={ }^{2 n} C_n$</p>
<p>$ (x+\frac{1}{x})^{2n} \xrightarrow[\text{coefficient}]{\text{middle}} { }^{2 n} C_n $</p>
<p>$\begin{aligned} & ={ }^{2 n} C_0 x^{2 n}+{ }^{2 n} C_1 x^{2 n-1} \cdot \frac{1}{x}+\cdots \cdot \\ & { }^{2 n} C_n \cdot  x^n(\frac{1}{x})^n+\cdots \cdot{ }^{2 n} C_{n-1} x \cdot \frac{1}{x^{2 n-1}}+{ }^{2 n} C_{2 n} \cdot \frac{1}{x^{2 n}} \\ & =(x+\frac{1}{x})^n(\frac{1}{x}+x)^n \\ & =({ }^n c_0 x^n+{ }^n c_1 x^{n-1} \cdot \frac{1}{x}+{ }^n c_2 x^{n-2} \cdot \frac{1}{x^2}+\cdots \cdot{ }^n c_n \frac{1}{x^n}) \\ & ({ }^n C_0 \frac{1}{x^n}+{ }^n C_1 \frac{1}{x^{n-1}} \cdot x+{ }^n C_2 \frac{1}{x^{n-2}} \cdot x^2+\cdots n^n C_n x^n)\\ & \end{aligned}$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-09-chapter-02-binomial-theorem-L-4-9-fpmf-zalzjw-6.jpg"></p>

### <Success>Result 4</Success>
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<p>$\begin{aligned} & C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n \\ & ={ }^{2 n} C_{r+n} =\frac{2 n !}{(r+n) !(n-r) !} \\ & \end{aligned}$</p>
<p><strong>Proof:</strong><br>
$r = 0 \longrightarrow C_0^2+C_1^2+C_2^2+\cdots C_n^2={ }^{2 n} C_n=\frac{2 n !}{n ! n !}$</p>
<p>$$
(x+y)^{2 n} \rightarrow{ }^{2 n} C_n x^n y^n \quad { }^{2 n} C_{n+1} x^{n-1} y^{n+1} { }^{2 n}\quad C_n+r{x^{n-r} y^{n+r}}
$$</p>
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### <Success>Problem</Success>
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<p>What is the coefficient of $x^{n-r} \cdot y^{n+r}$ in $(x+y)^n \cdot(y+x)^n$?</p>
<p><strong>Solution:</strong></p>
<p>$\begin{aligned}
& \rightarrow(x+y)^n \cdot(y+x)^n \\ & { }^n C_r x^{n-r} \cdot y^r{ }^n C_0 y^n\rightarrow{ }^n C_r{ }^n C_0 \\ & { }^n C_{r+1} x^{n-r-1} \cdot y^{r+1}{ }^n C_1 y^{n-1} x \rightarrow{ }^n C_{r+1}{ }^n C_1 \\ & { }^n C_{r+2}  x^{n-r-2} \cdot y^{r+2}. { }^nC_2 y^{n-2} x^2 \\ & \vdots \\ & { }^n C_n x^0 y^n \quad { }^n C_{n-r} y^r x^{n-r}\rightarrow{ }^n C_n{ }^n C_{n-r}\end{aligned}$</p>
<p>Thus, the coefficient of $(x+y)^n \cdot(y+x)^n$ is ${}^{2n}C_{n+r}.$</p>
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### <Success>Problem</Success>
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<p><strong>1.</strong> What is the coefficient of $x^7$ in $\left(a x^2+\frac{1}{b x}\right)^{11} \text{?} $<br>
<strong>2.</strong> What is the coefficient of $x^{-7}$ in $\left(a x-\frac{1}{b x}\right)^{11} \text{?} $<br>
<strong>3.</strong> What is the relation between $a, b$ if Ans $1=$ Ans 2</p>
<p>$\begin{aligned}&(a x^2+\frac{1}{b x})^{11}={ }^{11}C_0(a x^2)^{11}+{ }^{11}C_1(ax^2)^{10}(\frac{1}{b x})+{ }^{11}C_2(a x^2)^9(\frac{1}{b x})^2 \\ &{ }^{11}C_3^{(13)}+ { }^{11}C_4^{(10)}+{ }^{11}C_5(a x^2)^6(\frac{1}{b x})^5 \\ &\end{aligned}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p><strong>1)</strong>   ${ }^{11} C_5 a^6 / b^5$
$ \begin{aligned}
& \left(a x-\frac{1}{b x^2}\right)^{11}={ }^{1 1} C_0(a x)^{11}
+{ }^{11} C_1(a x)^{10}\left(\frac{1}{b x^2}\right)^{1}+{ }^{11} C_2() x^5-{ }^{11} C_3^{(2)} \\ & +{ }^{11} C_{4 x}^{(-1)}-{ }^{11} C_5 x^{-4}+{}^{11} C_6(a x)^5\left(\frac{1}{b x^2}\right)^6 \\ & \end{aligned} $</p>
<p><strong>2)</strong>  ${ }^{11} C_6 a^5 / b^6$</p>
<p>$ { }^{11} C_5=\frac{11 !}{5 !\cdot 6 !}={ }^{11} C_6 $</p>
<p>$ \frac{a^6}{b^5} $= $\frac{a^5}{b^6}$ $\Rightarrow a^6=\frac{a^5}{b} \ \Rightarrow a=\frac{1}{b} $</p>
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