### <Success>Infinite series lecture 2</Success> <div style='float: left; text-align:left width:100%;'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/014-0027.6.jpg"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/6e9e6b9b-08f8-4b38-7134-16de80c61e00/public"></p> </div>

### <Success>Recall</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Binomial Expressions</p> <p>$(1-r)^{-1}=1+r+r^2+r^3+\cdots $</p> <p>$\sum_{i=0}^{\infty} r^i \quad|r|<1 $</p> <p>$(1+r)^{-1}=1-r+r^2-r^3 $</p> <p>$\sum_{i=0}^{\infty}(-r)^i \ \ |r|<1 .$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/05c10856-f7a4-4aca-fd9f-37e637615e00/public"></p> </div>

### <Success>Question 1</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\text { What is }(0.8)^{-1} $</p> <p><strong>Solution:</strong></p> <p>$(0.8)^{-1}=\left(\frac{8}{10}\right)^{-1}=\left(\frac{10}{8}\right) $</p> <p>$=1.25$</p> <p><strong>Can we get the same answer by expanding its infinite sum?</strong></p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000001.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>We know $(0.8)^{-1}=(1-0.2)^{-1}$</p> <p>$(1-r)^{-1} \ \ |r| < 1 $</p> <p>$\because|0.2| < 1 .$</p> <p>By the series expansion.</p> <p>$\underbrace{1+0.2^2}_{=1.2}+ \underbrace{(0.2^2+0.2^3+ \cdots)} _{0.05 ?}$</p> <p>But the answer is $1.25$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/070-0357.6.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$0.2^2+0.2^3+0.2^4+\cdots \cdot$<br> $ = 0.2^2 (\underbrace{1+0.2+0.2^2+\cdots}_{\text {G.P. Series. }})$<br> $ = 0.2^2 \cdot \frac{1-r^n}{1-r}$<br> $= lim \ n \rightarrow \infty \ 0.2^2 \times \left(\frac{1-0.2^n}{1-0.2}\right)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fa3c6721-3bfb-4977-354e-59456f9f8b00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$0.2^2 \times \frac{1}{1-0.2} $</p> <p>$=0.2^2 \times \frac{1}{0.8}=\frac{0.04}{0.8} $</p> <p>$=\frac{1}{20}=0.05$</p> <p>$\therefore$ The whole sum is</p> <p>$1.2+0.05 $</p> <p>$= 1.25$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/105-0543.6.jpg"></p> </div>

### <Success>Question 2</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$(1+0.2)^{-1}$</p> <p><strong>Solution:</strong></p> <p>We knows the answer.</p> <p>$=(1 \cdot 2)^{-1}=\left(\frac{12}{10}\right)^{-1} $</p> <p>$=\frac{10}{12}=\frac{5}{6}$</p> <p>$=0.8333$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/123-0630.0.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$(1+r)^{-1}=1-r+r^2-r^3+ \cdots$<br> $\therefore (1+0.2)^{-1}$<br> $\underbrace{1-0.2}_{=0.8} + \underbrace{(0.2)^2-(0.2)^3+\cdots} _{=0.0333 \cdots}$</p> <p>$(0.2)^2(1-0.2 + 0.2^2 - \ldots)$<br> $\text { common ratio }, r=-0.2$<br> $\ =0.04\left(\frac{1}{1+0.2}\right)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/148-0787.2.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$0.04 \times \left(\frac{1}{1.2}\right)$</p> <p>$= \frac{4}{120}=\frac{1}{30}$</p> <p>$= 0.0333 .$</p> <p>$\therefore$ The answer is $0.8333 …$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/161-0853.2.jpg"></p> </div>

### <Success>Question</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Is it true for only $-1$ ?</p> <p>can we have similar expansion say for $-2,-3$?</p> <p>Or say some rational number say $\frac{1}{2}, \frac{2}{3}$ ?</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000002.png"></p> </div>

### <Success>Recall</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>We have studied Binomial Theorem for positive integer $n$.</p> <p>$(1+x)^n=1+{}^nC_1 x+{}^nC_2 x^2 +\cdots+{}^nC_{n-1} x^{n-1} + x^n$</p> <p><strong>a.</strong> Number of terms in finite.</p> <p><strong>b.</strong> We could use combination ${}^n C_r$ for $r=0,1, \cdot\cdot\cdot n.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2dda0919-cc58-482b-43f9-abf7a8609c00/public"></p> </div>

### <Success>Remark</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>The problem with negative index is</p> <p>i) The number of terms is infinite.</p> <p>ii) We cannot use ${}^{-n}C_r $ as This is not defined.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0a92b223-d0e4-4dbe-2cf1-08beb60cad00/public"></p> </div>

### <Success>$\text{Expansion of} \ (1-r)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us first see what is $(1-r)^{-2} .|r|<1$</p> <p>Our aim is to find out the coefficients of the expansion.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000003.png"></p> </div>

### <Success>$\text{Expansion of} \ (1-r)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us assume</p> <p>$(1-r)^{-2}=a_0+a_1 r+a_2 r^2+\cdots$</p> <p>We knows that $(1-r)^{-2}=(1-r)^{-1} \times (1-r)^{-1}$</p> <p>$\left(1+r+r^2+\cdots \cdot\right) \times \left(1+r+r^2+\cdots\right)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/247-1299.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \ (1-r)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\left(1+r+r^2+\cdots\right) \times \left(1+r+r^2+\cdots\right)$</p> <p>coeff of $r^0=1 \quad \therefore a_0=1$</p> <p>Coeff of $r=(1+1)=2 \quad \therefore a_1=2$</p> <p>coeff of $r^2=(1+1+1)=3 \quad \therefore a_2=3$</p> <p>coeff of $r^3=4 \quad \therefore a_3=4$</p> <p>In ganeral coeff of $r^k={k+1}$</p> <p>$\therefore (1-r)^{-2}$</p> <p>$=1+2 r+3 r^2+4 r^3+ \cdots$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/280-1530.0.jpg"></p> </div>

### <Success>$\text{Expansion of} \ (1-r)^{-2}$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>From here we can easily see that:</p> <p>$(1+r)^{-2}=(1-(-r))^{-2} $</p> <p>$=1-2 r+3 r^2-4 r^3+\cdots $</p> <p>$\text { Expand }(1+r)^{-2} \text { as } $</p> <p>$(1+r)^{-1} \times (1+r)^{-1}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000004.png"></p> </div>

### <Success>Question 3</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>what in $(1-r)^{-3}?$</p> <p><strong>Solution:</strong> Again as before we assume</p> <p>$(1-r)^{-3}=b_0+b_1 r+b_2 r^2+b_3 r^3 + \cdot \cdot \cdot$</p> <p>And we try to find the values of $b_0, b_1, b_2 \ldots$. etc.</p> <p>We will write</p> <p>$(1-r)^{-3}=(1-r)^{-2} \times (1-r)^{-1}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/315-1734.0.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ (1-r)^{-3}=b_0+b_1 r+b_2 r^2+\cdots \ (1-r)^{-2} \times (1-r)^{-1}= \ \left(1+2 r+3 r^2+4 r^3+\cdots\right) \ \ \times \left(1+ r+r^2+\cdots\right) \ \text { coeff of } r^0=1 \quad \therefore b_0=1 \ \text { coess of } r=(1+2) \therefore b_1=3 \ \text { coesf of } r^2=(1+2+3)=6 \quad \therefore b_2=6 \ \text { coesf } of r^3=(1+2+3+4)=b_3=10 $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/343-1918.8.jpg"></p> </div

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore(1-r)^{-3}$</p> <p>$=1+3 r+6 r^2+10 r^3+\ldots .$</p> <p>The general term: coeff of $r^k$ in.</p> <p>$\text { 1. }(k+1)+1 \cdot k+1 \cdot(k-1)+\cdots+1 \cdot 1$</p> <p>$=1 \cdot(1+2+\cdots+k+1) $</p> <p>$=\frac{(k+1)(k+2)}{2}=b_k= \ \text{Coeff of} \ r^k $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/362-2073.6.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\begin{aligned} \therefore & (1-r)^{-3} \\ & =\sum_{k=0}^{\infty} \frac{(k+1)(k+2)}{2} r^k \end{aligned}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/338702f2-8040-469e-face-eff318c5aa00/public"></p> </div>

### <Success>Notes</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Obviously, it is not possible to compute these terms for arbitrarily large $n.$ We have the formula:</p> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/385-2155.2.jpg"></p> </div>

### <Success>Computation of the formula</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>For positive integral $n$. we have coefficient of</p> <p>$r^k={}^n C_k \text {. }$ not valid for $-n$. we write ${}^nC_r$ as</p> <p>$\frac{n !}{r !(n-r) !}=\frac{r(n-1) \cdots(n-r-1)}{r !}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7496d35f-7921-4f8b-2d00-69dd6bad5c00/public"></p> </div>

### <Success>Computation of the formula</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us denote by $c(-n, k)$ the term $\frac{(-n)(-n-1)(-n-2) \cdots(-n-k+1)}{k !}$</p> <p>We shall see that $c(-n, k)$ in the coeff of $r^k$ in the expension of $(1+x)^{-x} \quad|x|<1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a82c662d-37a4-4124-2933-e2f071a8e200/public"></p> </div>

### <Success>Verification of the formula</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Consider</p> <p>$(1-r)^{-2}$ coeff of $-r=\frac{-2}{1!}=-2$</p> <p>coeff of $(-r)^2=\frac{-\dot{2}(-3)}{2!}=3$</p> <p>$\operatorname{coeff}(-r)^3=\frac{-2(-3 !(-4)}{3 !}=-4$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/443-2430.0.jpg"></p> </div>

### <Success>Verification of the formula</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ The expansion is:</p> <p>$1+(-2)(-r)+3(-r)^2+(-4)(-r)^3 + \cdot \cdot $</p> <p>$=1+2 r+3 r^2+4 r^3 + \cdot \cdot $</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/03c5febd-c9b9-44fb-2c0a-646189e5fb00/public"></p> </div>

### <Success>Verification</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us verify for $(1+r)^{-3}$ coeff $r=\frac{-3}{1!}=-3$</p> <p>coeff $r^2=\frac{-3(-4)}{2 !}=6$</p> <p>coeff $r^3=\frac{-3(-4)(-5)}{3 !}=-10$</p> <p>$(1+r)^{-3}$</p> <p>$1-3 r+6 r^2-10 r^3+\cdots$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/458-2584.8.jpg"></p> </div>

### <Success>Binomial expansion with fractional index</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>what is $(1-x)^{\frac{1}{2}}$ or $(1-x)^{-\frac{1}{2}}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000005.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Consider $(1-x)^{-\frac{1}{2}}$</p> <p>we know that</p> <p>$(1-x)^{-1}=(1-x)^{-\frac{1}{2}} \times (1-x)^{-\frac{1}{2}} |x|<1 $</p> <p>$ 1+x+x^2+x^3+\ldots $</p> <p>Let us assume</p> <p>$(1-x)^{-\frac{1}{2}}=a_0+a_1 x+a_2 x^2$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/478-2706.0.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ (1-x)^{-\frac{1}{2}} \times (1-x)^{-\frac{1}{2}} $</p> <p>$=\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right)$</p> <p>$\quad \times \left(a_0+a_1 x+a_2 x^2+a_3 x^3\right) = \ {1+x+x^2+x^3+\cdots} $</p> <p>${\therefore a_0^2= 1 \therefore a_0 = \pm1}$</p> <p>But we take the positive value</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ad90dbf0-6834-4d7a-ddfa-b7ab8a48e500/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore G_0=1$</p> <p>$\text { coeff of } x=G_0 G_1+G_1 G_0 $</p> <p>$ =2\cdot G_0 \cdot G_1 =2 G_1=1 $</p> <p>$\operatorname{from}(1-x)^{-1}$</p> <p>$2 G_1=1 \quad \therefore G_1=\frac{1}{2} $</p> <p>$\text { coeff } x^2=G_0 G_2+G_1^2+G_2 G_0 =1$</p> <p>$2 G_0 G_2+G_1^2=1 $</p> <p>$2 G_0 G_2=1-\frac{1}{4}=\frac{3}{4} $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/510-2913.6.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore 2G_2=\frac{3}{4} \quad \therefore G_2=\frac{3}{8} $</p> <p>$\therefore$ we get that</p> <p>$ (1-x)^{-\frac{1}{2}}=1+\frac{1}{2} x+\frac{3}{8} x^2+ \cdot \cdot \cdot $</p> <p>We shall see whether we get the same</p> <p>by using the expansion the way we did for negative integer</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000006.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>For negative integer we have seen that coeff of $x^k$ in</p> <p>$ (1-x)^{-n}=\frac{(-n)(-n-1)–(-n-k+1)}{k !} $</p> <p>Let us apply the same for $\frac{p}{a}$, in particular here for $-\frac{1}{2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000007.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>By using similar expansion we get:</p> <p>$\quad(1-x)^{-\frac{1}{2}}=$</p> <p>$1-\left(-\frac{1}{2}\right) x+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2 !}(x)^2 $</p> <p>$=1+\frac{1}{2} x+\frac{3}{8} x^2+\cdots $</p> <p>$G_0=1 $</p> <p>$G_1=\frac{1}{2}$</p> <p>$G_2=\frac {3}{8}$</p> <p>This we get the same answer</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-14-infinite-series-l-2_5-gmb6dxjcalc/img/000008.png"></p> </div>

### <Success>Thank you</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> </div> <div style='float: right; width:0%;'>