mathematics-class-11-unit-09-chapter-01-binomial-expansions-L-1-9-thzfnuzc9jc

### <Success>Question 1</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>Which one among $101^{50}$ and $99^{50}+100^{50}$ is bigger?</p>
<p><strong>Solution:</strong></p>
<p>$101^{50} = (100+1)^{50} $</p>
<p>$99^{50}=(100-1)^{50} $</p>
<p>$101^{50}-99^{50}=(100+1)^{50}-(100-1)^{50} $</p>
<p>$=\sum_{k=0}^{50}{ }^{50} C_k(100)^{50-k}-\sum_{k=0}^{50}{ }^{50} C_k(100)(-1) $</p>
<p>$=2 \sum_{k=0}^{50}{ }^{50} C_k(100)^{50-k} \ \ (k \ \text{odd})$</p>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$ 2({ }^{50} C_1 100^{49}+{ }^{50} C_3 100^{47}+ …. +{ }^{50} C_{49} 100) $</p>
<p>$\underbrace{50}  \ \ \{\because { }^{50} C_1 =50 \}  $</p>
<p>$=100^{50}+2^{50} C_3 100^{47}+\cdots+2^{50} C_{49} 100 $</p>
<p>$101^{50}-99^{50}>100^{50} $</p>
<p>$\Rightarrow 101^{50}>100^{50}+99^{50} $</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-6.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-01-binomial-expansions-L-1-9-thzfnuzc9jc-6a.jpg"></p>

### <Success>Question 2</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>If for a positive integer $n$, the coefficients of $2 n d, 3 r d$ and $4 t h$ term in the expansion of $(1+x)^n$ are in arithmetic progression, then what is $n$ ?</p>
<p><strong>Solution:</strong></p>
<p>$(1+x)^n$=</p>
<p>${ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\cdots+{ }^n C_n x^n $</p>
<p>$\text { 2nd term } \longrightarrow{ }^n C_1=(n) $</p>
<p>$\text { 3rd term } \longrightarrow{ }^n C_2=\left(\frac{n(n-1)}{2}\right) $</p>
<p>$\text {4th term } \rightarrow{ }^n C_3=\left(\frac{n(n-1)(n-2)}{6}\right. $</p>
<p>$\frac{n+\frac{n(n-1)(n-2)}{6}}{2}=\frac{n(n-1)}{2}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$ 6+(n-1)(n-2)=6 n-6 $</p>
<p>$\Rightarrow 6+n^2-n-2 n+2=6 n-6 $</p>
<p>$\Rightarrow n^2-9 n+14=0 $</p>
<p>$n=\frac{9 \pm \sqrt{81-56}}{2} =\frac{9 \pm 5}{2} $</p>
<p>$n= \underbrace{2}_{\text{Not possible}} \text { or } 7 $</p>
<p>$\biggl\{ (1+x)^n \ \ \text { there are total $ n+1$ terms } \\ \text{If} \ n=2 , \ \text{we get 3 terms}\biggl \}$</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-7.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-01-binomial-expansions-L-1-9-thzfnuzc9jc-7a.jpg"></p>

### <Success>Question 4</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>The expression $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$ is a polynomial of degree</p>
<p>$(1) 5 \quad $  $(2) 6 \quad $ $(3) 7 \quad $ $ (4) 8$</p>
<p><strong>Solution:</strong></p>
<p>$\left(x+\sqrt{x^3-1}\right)^5=\sum_{k=0}^5 5 C_k x^{5-k}\left(\sqrt{x^3-1}\right)^k  $</p>
<p>$\left(x-\sqrt{x^3-1}\right)^5=\sum_{k=0}^5 5 C_k x^{5-k}(-1)^k\left(\sqrt{x^3-1}\right)^k $</p>
<p>$\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5=2 \sum_{k=0}^5 C_k x^{5-k}\left(x^3-1\right)^{k / 2}$ k even</p>
<p>$[k=0 \quad x^5  \quad \text {1 degree 5} ] $</p>
<p>$[k=2 \quad x^3\left(x^3-1\right) \text { degree 6} ] $</p>
<p>$[k=4 \quad x(x^3-1)^2 \text{ degree 7} ] $</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-9.jpg"></p>

### <Success>Question 5</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>The term independent of $x$ in the expansion of $(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1} -  \frac{x-1}{x-x^{1 / 2}})^{10}$ is</p>
<p>$(1) 4$</p>
<p>$(2) 120$</p>
<p>$(3) 210$</p>
<p>$(4) 310$</p>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$ \frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}=\frac{\left(x^{1 / 3}\right)^3+1}{x^{2 / 3}-x^{1 / 3}+1}=\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)} $</p>
<p>$\frac{x-1}{x-x^{1 / 2}}=\frac{\left(x^{1 / 2}\right)^2-1}{x^{1 / 2}\left(x^{1 / 2}-1\right)}=\frac{\left(x^{1 / 2}+1\right)\left(x^{1 / 2}-1\right)}{x^{1 / 2}\left(x^{1 / 2}-1\right)}=\frac{x^{1 / 2}+1}{x^{1 / 2}} $</p>
<p>$=1+x^{-1 / 2}$</p>
<p>$\therefore (\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1} -  \frac{x-1}{x-x^{1 / 2}})^{10} =(x^{1 / 3}+1-1-x^{-1 / 2})^{10} $</p>
<p>$=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$</p>
<p>$=\sum_{k=0}^{10}{ }^{10} C_k\left(x^{1 / 3}\right)^{10-k}(-1)^k x^{-k / 2} $</p>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$=\sum_{k=0}^{10}(-1)^k  \quad { }^{10} C_k x^{(20-5 k) / 6} $</p>
<p>$\frac{10-k}{3}-\frac{k}{2} $</p>
<p>$=\frac{20-2 k-3 k}{6} $</p>
<p>$=\frac{20-5 k}{6} $</p>
<p>$\frac{20-5 k}{6}=0 $</p>
<p>$\Rightarrow k=4 $</p>
<p>$(-1)^4$ ${ }^{10} C_4 $</p>
<p>$=1 \cdot \frac{10 \cdot 9 \cdot 8 \cdot 7}{24}=210$</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-10.jpg"></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-01-binomial-expansions-L-1-9-thzfnuzc9jc-10a.jpg"></p>
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### <Success>Question 6</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>The value of the sum ${ }^{50} C_4+\sum_{r=1}^6{ }^{56-r} C_3$ is</p>
<p>$(1) { }^{55} C_4 $</p>
<p>$(1) { }^{55} C_3 $</p>
<p>$(1) { }^{56} C_3 $</p>
<p>$(1) { }^{56} C_4 $</p>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$n \in \mathbb{N}$, $\ \ 0 \leqslant r < n$,</p>
<p>We have,</p>
<p>${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$</p>
<p>${ }^{50} C_4+{ }^{50} C_3+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$</p>
<p>$={ }^{51} C_4+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$</p>
<p>$={ }^{52} C_4+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$</p>
<p>$={ }^{53} C_4+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$</p>
<p>$={ }^{54} C_4+{ }^{54} C_3+{ }^{55} C_3$</p>
<p>$={ }^{55} C_4+{ }^{55} C_3$</p>
<p>$={ }^{56} C_4$</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-11.jpg"></p>

### <Success>Question 8</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>The value of ${ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-\cdots+{ }^{20} C_{10}$ is</p>
<p>(1) 0</p>
<p>(2) ${ }^{20} C_{10}$</p>
<p>(3) $-{ }^{20} C_{10}$</p>
<p>(4) $\frac{{ }^{20} C_{10}}{2}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$ Y={ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-\cdots+{ }^{20} C_{10} $</p>
<p>$ (1-x)^{20}={ }^{20} C_0-{ }^{20} C_1 x+{ }^{20} C_2 x^2-\cdots $</p>
<p>$ +{ }^{20} C_{10} x^{10}-20 C_{11} x^{11}+\cdots+{ }^{20} C_{20} x^{20} $</p>
<p>$\text { Put } x  =1 $</p>
<p>$0  =Y-{ }^{20} C_{11}+{ }^{20} C_{12}-\cdots+{ }^{20} C_{20}$</p>
<p>$Y={ }^{20} C_9-{ }^{20} C_8+{ }^{20} C_7-\cdots-{ }^{20} C_0 +{ }^{20} C_{10}-{ }^{20} C_{10} $</p>
<p>= $ -{\left({ }^{20} C_{10}-{ }^{20} C_9+{ }^{20} C_8-\cdots+{ }^{20} C_0\right)}{+{ }^{20} C_{10}} $</p>
<p>$2 Y={ }^{20} C_{10} $</p>
<p>$ Y=\frac{1}{2}{ }^{20} C_{10} $</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-14.jpg"></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-01-binomial-expansions-L-1-9-thzfnuzc9jc-14a.jpg"></p>
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### <Success>Question 9</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>The value of the sum
${ }^{30} C_0{ }^{30} C_{10}-{ }^{30} C_1{ }^{30} C_{11}+{ }^{30} C_2{ }^{30} C_{12}-\cdots+{ }^{30} C_{20}{ }^{30} C_{30}$ is</p>
<p>(1) ${ }^{30} C_{10}$</p>
<p>(2) ${ }^{30} C_{15}$</p>
<p>(3) ${ }^{60} C_{30}$</p>
<p>(4) ${ }^{31} C_{10}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess-2"><Success>Solution</Success></h3>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>${ }^{30} C_0{ }^{30} C_{20}-{ }^{30} C_1{ }^{30} C_{19}+{ }^{30} C_2{ }^{30} C_{18}-\cdots+{ }^{30} C_{20}{ }^{30} C_0$</p>
<p>co-efficient of $x^{20}$ in $(1+x)^{30}(1-x)^{30}$</p>
<p>$(1+x)^{30}(1-x)^{30} $</p>
<p>$=\left(1-x^2\right)^{30} $</p>
<p>$=\sum_{k=0}^{30}{ }^{30} C_k(-1)^k x^{2 k} $</p>
<p>$\text { co-efficient of } x^{20}={ }^{30} C_{10}(-1)^{10} $ $={ }^{30} C_{10}$</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-01-binomial-expansions-L-1-9-thzfnuzc9jc-15a.jpg"></p>
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### <Success>Question 10</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>The sum of the coefficients of the integral powers of $x$ in the binomial expansion of $(1-2 \sqrt{x})^{50}$ is</p>
<p>(1) $\frac{3^{50}-1}{2}$</p>
<p>(2) $\frac{3^{50}+1}{2}$</p>
<p>(3) $\frac{2^{50}+1}{2}$</p>
<p>(4) $\frac{3^{50}}{2}$</p>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:25px;'>
<p>$(1-2 \sqrt{x})^{50} = \sum_{k=0}^{50}{ }^{50} C_k(-1)^k 2^k x{ }^{k / 2} $</p>
<p>${ }^{50} C_0+{ }^{50} C_2 2^2+{ }^{50} C_4 2^4+\cdots+{ }^{50} C_{50} 2^{50}$</p>
<p>$ (1+2 x)^{50}+(1-2 x)^{50} $</p>
<p>$=  \sum_{k=0}^{50}{ }^{50} C_k 2^k x^k+\sum_{k=0}^{50}{ }^{50} C_k(-1)^k 2^k x^k $</p>
<p>$=  2 \sum_{k=0}^{50}{ }^{50} C_k 2^k x^k \ \ k= \text{even}$</p>
<p>Putting $x=1$,</p>
<p>$ { }^{50} C_0+{ }^{50} C_2 2^2+{ }^{50} C_4 2^4+\cdots+{ }^{50} C_{50} 2^{50} $</p>
<p>$=\frac{1}{2}\left\{3^{50}+(-1)^{50}\right\}$</p>
<p>$ =\frac{3^{50}+1}{2} $</p>
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<p>======</p>
<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3>
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