mathematics-class-11-unit-09-chapter-02-binomial-expansions-l-2-9-6iadeej0uzc

### <Success>Question 12</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>The coefficient of $t^{24}$ in the expansion of $(1+t^2)^{12}(1+t^{12})(1+t^{24})$ is</p>
<p>(1) ${ }^{12} C_6 $</p>
<p>(2)  ${ }^{12} C_6 + 1 $</p>
<p>(3)  ${ }^{12} C_6 + 2 $</p>
<p>(4)  ${ }^{12} C_6 + 3 $</p>
</div>
<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>$(1+t^2)^{12}.(1+t^{12}).(1+t^{24}) $</p>
<p>$ =\underset{k=0} {\overset {12}\sum} .{ }^{12} C_k .t^{24-2 k} (1+t^{12})(1+t^{24}) $</p>
<p>$ =\underset {k=0}{\overset {12}\sum} .{ }^{12} C_k (t^{24-2 k}+t^{36-2 k}+t^{48-2 k}+t^{60-2 k})$</p>
<p>$ { }^{12} C_0 + { }^{12} C_6 + { }^{12} C_{12} $</p>
<p>$ = 1 + { }^{12} C_6 + 1 = 2 + { }^{12} C_6 $</p>
</div>
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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-expansions-L-2-9-6Iadeej0uzc-7.jpg"></p>

### <Success>Question 14</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>If for some integers $n, r$ such that $0 \leq r<n$, we have for some real number $k,{ }^{n-1} C_r = (k^2-3){ }^n C_{r+1}$. Then $k$ can lie in which of the following intervals?</p>
<p>( 1)($-\infty,-2]$</p>
<p>(2) $[2, \infty)$</p>
<p>(3) $[-\sqrt{3}, \sqrt{3}]$</p>
<p>(4) $(\sqrt{3}, 2]$</p>
</div>
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### <Success>Solution</Success>
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<p>$ { }^{n-1} C_r=(k^2-3) { }^n C_{r+1} $</p>
<p>$\Rightarrow  \frac{(n-1) !}{r !(n-1-r) !}=(k^2-3) \frac{n !}{(r+1) !(n-k-1) !}$</p>
<p>$\Rightarrow  \frac{k+1}{n}=k^2-3, \quad \frac{1}{n} \leq \frac{k+1}{n} \leq 1$</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-expansions-L-2-9-6Iadeej0uzc-10.jpg"></p>

### <Success>Question 15</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>The remainder of $8^{2020} - 62^{2021}$ when divided by 9 is</p>
<p>(1) 2</p>
<p>(2) 7</p>
<p>(3) 8</p>
<p>(4) 0</p>
</div>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>$ 8 = 9-1 $</p>
<p>$ 62 = 63 - 1    \quad   [ 63 = 7 \times 9] $</p>
<p>= $8^{2020} - 62^{2021}$</p>
<p>= $(9-1)^{2020} - (63-1)^{2021}$</p>
<p>= $\underset {k=0}{\overset {2020}\sum} { }^{2020}C_k. 9^{2020 - k} (-)^k - \underset{k=0} {\overset {2021} \sum}. { }^{2021} C_r. 63^{2021-r} (-1)^r $</p>
<p>= ${ }^{2020} C_{2020} - { }^{2021} C_{2021} (-1)$</p>
<p>= 2</p>
</div>
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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-expansions-L-2-9-6Iadeej0uzc-12.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-12a.jpg"></p>

### <Success>Question 16</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>For a,b$\neq$0, the sum of the $5^{th}$ and the $6^{th}$ terms in the expansion of $(a-b)^n$ is zero for an integer $n\geq5$. Then the ratio a/b is.</p>
<p>(1) $\frac {n-5}{6}$</p>
<p>(2) $\frac {n-4}{5}$</p>
<p>(3) $\frac {5}{n-4}$</p>
<p>(4) $\frac {6}{n-5}$</p>
</div>
<p>======</p>
<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
<div style='float: left;text-align:left; width:100%;font-size:27px;'>
<p>$\begin{aligned}& (a-b)^n=\sum_{k=0}^n{ }^n C_k a^{n-k}(-1)^k b^k \\ & T_5={ }^n C_4 a^{n-4} b^4 \\ & T_6=-{ }^n C_5 a^{n-5} b^5\end{aligned}$</p>
<p>$ { }^n C_4 a^{n-4} b^4-{ }^n C_5 a^{n-5} b^5=0 $</p>
<p>$ \Rightarrow{ }^n C_4 a^{n-5} b^4 {a-\frac{(n-4)b}{5}} =0 $</p>
<p>$ \text { As } a, b \neq 0, $</p>
<p>$ a-\frac{(n-4) b}{5}=0 $</p>
<p>$ \Rightarrow \frac{a}{b}=\frac{n-4}{5} $</p>
</div>
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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-expansions-L-2-9-6Iadeej0uzc-13.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-13a.jpg"></p>

### <Success>Question 17</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>For a positive integer $n$, the coefficients of three consecutive terms in the expansion of $(1+x)^{n+5}$ are having ratio $5: 10: 14$. Then what is the value of $n$ ?</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>$ (1+x)^{n+5}=\sum_{t=0}^{n+5} n+5 C_t x^t $</p>
<p>$ { }^{n+5} C_{r-1},{ }^{n+5} C_r,{ }^{n+5} C_{r+1} \text { for some }  0<r<n+5 $</p>
<p>$ \biggr[{ }^{n+5} C_{r-1} = 5k \biggr], \ \biggr[{ }^{n+5} C_r = 10k \biggr], \ \biggr[{ }^{n+5} C_{r+1} = 14k \biggr] $</p>
<p>for some $k$</p>
<p>$ \frac{(n+5) !}{(r-1) !(n+6-r) !}=5 k $</p>
<p>$ \frac{(n+5) !}{r !(n+5-r) !}=10 k $</p>
<p>$ \frac{(n+5) !}{(r+1) !(n+4-r) !}=14 k $</p>
</div>
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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-expansions-L-2-9-6Iadeej0uzc-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-02-binomial-expansions-L-2-9-6Iadeej0uzc-14a.jpg">

### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:27px;'>
<p>$\frac{(1+x)^{50}-(1+x)^2}{x}+{(1+m x)^{50}}$</p>
<p>co-efficient of $x^2$ in the given sum  $={}^{50}C_3+m^2 \quad {}^{50}C_2$</p>
<p>${}^{50} C_3+m^{2} \quad {}^{50} C_2={ }^{51}C_3(3 n+1)$</p>
<p>$\Rightarrow(3 n+1)=\frac{{}^{50} C_3}{{}^{51} C_3}+m^2 \frac{{}^{50} C_2}{{}^{51} C_3}$</p>
<p>$\begin{aligned} & 3 n+1=\frac{48}{51}+m^2 \frac{3}{51} \\ & \Rightarrow 153 n+51 =48+3 m^2 \\ & \Rightarrow \quad 3 m^2-3=153 n \\ & m^2-1=51 n  \\ & m=1, n=0 \ & \end{aligned}$</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-expansions-L-2-9-6Iadeej0uzc-17.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-17a.jpg"></p>

### <Success>Question 19</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>For integers $m, n$ with $n \geq m$, show that</p>
<p>${ }^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m={ }^{n+1} C_{m+1}.$</p>
<p>Hence prove that</p>
<p>$ {}^nC_m+2 {}^{n-1}C_m+3 {}^{n-2}C_m+\cdots+(n-m+1) \quad{}^mC_m={}^{n+2}C_{m+2}$</p>
</div>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>$ { }^m C_m+{ }^{m+1} C_m+{ }^{m+2} C_m+\cdots+{ }^{n-1} C_m+{ }^n C_m$</p>
<p>$[{ }^m C_m+{ }^{m+1} C_m = { }^{m+1} C_{m+1}]$</p>
<p>$ ={}^{m+2} C_{m+1}+{ }^{m+2} C_m+ \ldots+{ }^{n-1} C_m+{ }^n C_m$</p>
<p>$[{}^{m+2} C_{m+1}+{ }^{m+2} C_m  = {}^{m+3}C_{m+1}]$</p>
<p>${ }^{n-1}{C_{m+1}}{ }^{n-1} C_m+{ }^n C_m$</p>
<p>$={ }^{n} C_{m+1}+{ }^n C_m$</p>
<p>$={ }^{n+1} C_{m+1}$</p>
</div>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>$\begin{aligned}& { }^n C_m+2^{n-1} C_m+3^{n-2} C_m+\cdots+(n-m+1)^m C_m \\& =({}^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m) \\& +\left({ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m\right) \\& +\cdots+\left({ }^{m+1} C_m+m C_m\right)+{ }^m C_m \\& ={ }^{n+1} C_{m+1}+{ }^n C_{m+1}+\cdots+{ }^{m+2} C_{m+1}+{ }^{m+1} C_{m+1} \\& ={ }^{n+2} C_{m+2} \text {. } \\&\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-expansions-L-2-9-6Iadeej0uzc-18.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-18a.jpg"></p>

### <Success>Question 20</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>Statement $1: \sum_{r=0}^n(r+1)^n C_r=(n+2) 2^{n-1}$.</p>
<p>Statement 2 : $\sum_{r=0}^n(r+1)^n C_r x^r=(1+x)^n+n x(1+x)^{n-1}$.</p>
<ul>
<li>Statement 1 is false, Statement 2 is true.</li>
<li>Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.</li>
<li>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.</li>
<li>Statement 1 is true, Statement 2 is false.</li>
</ul>
</div>
<p>======</p>
<h3 id="successsolutionsuccess-3"><Success>Solution</Success></h3>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>Put $x=1$ in statement 2</p>
<p>$\text { LHS }=\sum_{r=0}^n(r+1)^n C_r$</p>
<p>$\text { RHS }=2^{n}+n 2^{n-1}=2^{n-1}(2+n)$</p>
<p>$(1+x)^n=\sum_{r=0}^n{ }^n C_r x^r$</p>
<p>$x(1+x)^n=\sum_{r=0}^n{ }^n C_r x^{r+1}$</p>
<p>$(1+x)^n+n x(1+x)^{n-1}$</p>
<p>$=\sum_{r=0}^n(r+1)^n C_r x^r$</p>
<p>statement 2 is true.</p>
</div>
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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-expansions-L-2-9-6Iadeej0uzc-19.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-19a.jpg"></p>

### <Success>Question 21</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>Let $S_1=\sum_{j=1}^{10} j(j-1){ }^{10} C_j, S_2=\sum_{j=1}^{10} j^{10} C_j$ and $S_3=\sum_{j=1}^{10} j^2{ }^{10} C_j$</p>
<p>Statement 1 : $S_3=55 \times 2^9$</p>
<p>Statement 2 : $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.</p>
<ul>
<li>Statement 1 is false, Statement 2 is true.</li>
<li>Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.</li>
<li>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.</li>
<li>Statement 1 is true, Statement 2 is false.</li>
</ul>
</div>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:30px;'>
<p>$S_1+S_2=S_3$</p>
<p>$(1+x)^{10} =\sum_{j=0}^{10}{ }^{10} C_j x^j \\ 10(1+x)^9  =\sum_{j=0}^{10}{ }^{10} C_j j x^{j-1} $<br>
$\text { Put } x  =1$<br>
$\text { LHS }  =10 \cdot 2^9$<br>
$\text { RHS } =\sum_{j=1}^{10} j^{10} C_j=S_2 \\ S_2  =10 \cdot 2^9$</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-expansions-L-2-9-6Iadeej0uzc-20.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-02-binomial-expansions-L-2-9-6Iadeej0uzc-20a.jpg"></p>