### <Success>Binomial theorem and its applications</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-2.jpg"></p> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-2a.jpg"></p>

### <Success>Recall</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$(x+y)^n={ }^n c_0 x^n+{ }^n C_1 x^{n-1} y + { }^n C_2 x^{n-2} y^2+{ }^n C_3 x^{n-3} y^3 + \cdots +{ }^n C_{n-1} x y^{n-1}+{ }^n C_n y^n $</p> <p>$\biggl [{ }^n C_r=\frac{n !}{r !(n-r) !}\biggl]$</p> </div> <div style='float: right; width:1%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-3.jpg"></p>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>In $(a - b)^n \ (n \geqslant 5),$ the sum of $5^{\text{th}}$ and $6^\text{{th}}$ terms is $0.$ Then find the value of $\frac{a}{b}.$</p> <p><strong>Solution:</strong></p> <p>$ (a-b)^n={ }^n C_0 a^n- { }^n C_1 a^{n-1} b+\quad+ \underbrace{{ }^n C_4 a^{n-4} b^4} _{5^{\text{th}} \ \text{term}} - \underbrace{{ }^n C _5 a^{n-5} b^5} _{6^{\text{th}} \ \text{term}} $<br> Now, according to the question,</p> <p>${ }^n C_4 a^{n-4} b^4={ }^n C_5 a^{n-5} b^5 $</p> <p>$\frac{n !}{4 ! n-4 !} a=\frac{n !}{5 ! n-5 !} b$</p> <p>$a / b=\frac{n-4}{5}$</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-7.jpg">

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:24px;'> <p>$\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\cdots a_{2 n} x^{2 n}$</p> <p><strong>1)</strong> $a_0+a_1+a_2+\ldots+a_{2 n}=$ ?</p> <p><strong>2)</strong> $a_0-a_1+a_2-a_3+a_4-\ldots+a_{2 n}=$ ?</p> <p><strong>3)</strong> $a_0^2-a_1^2+a_2^2-a_3^2+\cdots+a_{2 n}^2=$ ?</p> <p><strong>Solution:</strong></p> <p><strong>1)</strong> $a_0+a_1+a_2+\ldots+a_{2 n}= 3^n$<br> <strong>2)</strong> $a_0-a_1+a_2-a_3+a_4-\ldots+a_{2 n}= 1$<br> <strong>3)</strong> $(1+x+x^2)^n (1-\frac{1}{x}+\frac{1}{x^2})^n $</p> <p>$a_0+a_1 x+a_2 x^2+\cdots +a_{2 n} x^{2 n}$</p> <p>$a_0-\frac{a_1}{x}+\frac{a_2}{x^2}-\quad+a_{2 n} / x^{2 n}$</p> <p>term independent of $x$ is $a_0^2-a_1^2+a_2^2-a_3^2+\cdots \cdot .$</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-8.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <!--<div style='float: left;text-align:left; width:100%;font-size:30px;'>--> <!--$\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\cdots a_{2 n} x^{2 n}$--> <!--1) $a_0+a_1+a_2+\ldots+a_{2 n}=$ ? $3^n$--> <!--2) $a_0-a_1+a_2-a_3+a_4-\ldots+a_{2 n}=$ ? 1--> <!--3) $a_0^2-a_1^2+a_2^2-a_3^2+\cdots+a_{2 n}^2=$ ?--> <!--$(1+x+x^2)^n (1-\frac{1}{x}+\frac{1}{x^2})^n $--> <!--$a_0+a_1 x+a_2 x^2+\cdots +a_{2 n} x^{2 n}$ --> <!--$a_0-\frac{a_1}{x}+\frac{a_2}{x^2}-\quad+a_{2 n} / x^{2 n}$--> <!--term independent of $x$ is--> <!--$--> <!--a_0^2-a_1^2+a_2^2-a_3^2+\cdots \cdot--> <!--$--> <!--</div>--> <p>$\left(1+x+x^2\right)^n \times\left(1-\frac{1}{x}+\frac{1}{x^2}\right)^n $</p> <p>$ = {\left[\left(1+x+x^2\right)\left(1-\frac{1}{x}+\frac{1}{x^2}\right)\right]^n }$</p> <p>$ = [1- \frac{1}{x} + \frac{1}{x^2} + x -1 + \frac{1}{x}+ x^2 - x +1]^n $</p> <p>$=\left[\frac{1}{x^2}+1+x^2\right]^n$</p> <p>$=\frac{1}{x^{2 n}}\left(1+x^2+x^4\right)^n $</p> <p>$=\frac{1}{x^{2 n}}\left(a_0+a_1 x^2+a_2 x^4+\cdots a_{n} x^{2 n} \cdots a_{2 n} x^{4 n}\right)$</p> <p>$\Rightarrow$ term independent of $x$ is $a_n$.</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-9.jpg"> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-9a.jpg"></p>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:27px;'> <p>$(1+x)^{14}$ if the $r^{\text {th }}, r+1^{\text {th }}, r+2^{\text {th }}$ terms are in Arithmetic Progression (AP). $r=$ ?</p> <p><strong>Solution:</strong></p> <p>$ r\text { th } \rightarrow 14-r+1 $ $\rightarrow \quad { }^{14} c_{r-1} \cdot x^{r-1} $</p> <p>$ r+1 \text { th } \rightarrow \quad{ }^{14} \mathrm{C}_r x^r $</p> <p>$ r+2 \text { th } \rightarrow \quad{ }^{14} c_{r+1} x^{r+1} $</p> <p>$ \frac{14 !}{(r-1) !(15-r) !} \cdot x^{r-1} +\frac{14 !}{(r+1) !(13-r) !} \cdot x^{r+1}=2 \frac{14 !}{(4-r) ! r !} x^r $<br> $\frac{1}{(14-r)(15-r)} +\frac{x^2}{r(r+1)}=\frac{2 x}{r(14-r)}$</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-10.jpg"> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-10a.jpg"></p>

### <Success>Remark</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$(1+x)^n=\frac{n !}{0 ! n !}+\frac{n !}{1 !(n-1) !} x +\frac{n !}{2 !(n-2) !} x^2+\cdots$</p> <p>$(1+x)^n= 1+n x+\frac{n(n-1)}{2} x^2+\frac{n(n-1)(n-2)}{3 !} x^3+ $</p> <p>$ \frac{n(n-1)(n-2)(n-3)}{4 !}+\cdots \cdot \ \ \ $ $|x| < 1$</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-12.jpg"></p>

### <Success>Recall: limit</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)=1 $</p> <p>$\lim _{x \rightarrow 0} e^x-1=\lim _{x \rightarrow 0} 1+x$</p> <p>$e =\lim _{x \rightarrow 0}(1+x)^{1 / x} $</p> <p>$=2 \cdot 7182818 \ldots $</p> <p>$ =\lim _{y \rightarrow \infty}(1+1 / y)^y$</p> <p>$e^x=\lim _{y \rightarrow \infty}(1+1 / y)^{x y}$</p> <p>$=\lim _{y \rightarrow \infty}\left[1+x y+\frac{x y(x y-1)}{2 !} \cdot \frac{1}{y^2} + \frac{x y(x y - 1)(x y-2)}{3 ! y^3} +\ldots \right] $</p> </div> <p>======</p> <h3 id="successrecall-limitsuccess"><Success>Recall: limit</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$ e^x=\lim _{y \rightarrow \infty}\left[1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !}+\cdots\right] $</p> <p>$ e^{-x}=\left[1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}+\cdots \cdot\right] $</p> <p>$ e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots $</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-13.jpg"></p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-13a.jpg"></p>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:28px;'> <p>Prove that $ C_1+2 C_2+3 C_3+\cdots+n C_n=n \cdot 2^{n-1} $</p> <p><strong>Solution:</strong><br> $ C_0+c_1+C_2+C_3+\cdots \quad C_n=2^n $</p> <p>$\frac{d}{d x} x^n=n x^{n-1}$</p> <p>$ (1+x)^n=C_0+C_1 x+C_2 x^2+\cdots \cdot C_n x$</p> <p>$ n(1+x)^{n-1}=C_1+2 C_2 x+3 C_3 x^2+\cdots+n C_n x^{n-1} $</p> <p>Plug in$ x=1 $</p> <p>$\Rightarrow n \cdot 2^{n-1}=C_1+2 C_2+3 C_3+\ldots+n C_n $</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-14.jpg"></p> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-14a.jpg"></p>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:26px;'> <p>$C_0+2 C_1+3 C_2+\cdots+(n+1) C_n=(n+2) 2^{n-1}$</p> <p><strong>Solution:</strong></p> <p>L.H.S $ = \left(C_0+C_1+C_2+\cdots C_n\right) +\left(C_1+2 C_2+3 C_3+\cdots+n C_n\right) \\ = 2^n +n \cdot 2^{n-1} $</p> <p>$=(n+2) 2^{n-1} $</p> <p>Other Way to prove,</p> <p>$C_0+2 C_1+3 c_2+\cdots+(n+1) C_n=k $</p> <p>$(n+1) C_n+n C_{n-1}+(n-1) C_{n-2}+\ldots+C_0=k $</p> <p>$(n+2) C_0+(n+2) C_1+(n+2) C_2+\cdots(n+2) C_n=2 k $</p> <p>$ k=\frac{n+2}{2}\left(C_0+C_1+C_2+\ldots+C_n\right)=\frac{n+2}{2} \cdot 2^n$</p> </div> <div style='float: right; width:0%;'> <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-03-binomial-theorem-L-5-9-4-08jfy-m6a-15.jpg"></p> </div> <p>======</p> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> </div>