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- Question 4 & 5 Exercise 5.2
- t \dfrac{3}{2}+\dfrac{3}{2}=\dfrac{18}{2}=9 \\ & \Rightarrow S_{\infty}=9\end{align} =====Question 5===== If ... \ldots \infty] \\ & =3+2 \cdot \dfrac{r}{1-r} \\ \Rightarrow S_{\infty}&=3+\dfrac{2 r}{1-r}\end{align} But we ... \therefore 3+\dfrac{2 r}{1-r}=\dfrac{44}{9} \\ & \Rightarrow \quad \dfrac{2 r}{1-r}=\dfrac{44}{9}-3=\dfrac{17}{9} \\ & \Rightarrow 18 r=17-17 r \\ & \Rightarrow 18 r+17 r=17 \\ &
- Question 5 Exercise 5.3
- text {terms } \\ & =\dfrac{6[2^{n-1}-1]}{2-1} \\ \Rightarrow a_n-a_1&=6 \cdot 2^{n-1}-6 \\ \Rightarrow a_n&=6 \cdot 2^{n-1}-6+a_1 \\ \Rightarrow a_n&=6 \cdot 2^{n-1}-6+3 \quad \because a_1=3\\ \Rightarrow a_n&=3(2^n-1)\end{align} Taking summation of the
- Question 2 Exercise 5.3
- }[6 n+8] \\ & =2 \cdot \dfrac{n-1}{2}[3 n+4] \\ \Rightarrow a_n-a_1&=(n-1)(3 n+4) \\ \Rightarrow a_n&=3 n^2+n-4+a_1 \\ \Rightarrow a_n&=3 n^2+n-4+4 \quad \because a_1=4 \\ \Rightarrow a_n&=3 n^2+n\end{align} Taking summation of the both sid
- Question 9 Review Exercise
- \ & =\dfrac{(n-1)(2 n+4)}{2} \\ & =(n-1)(n+2) \\ \Rightarrow a_n&=n^2+n-2+a_1 \\ \Rightarrow a_n&=n^2+n-2+3 \because a_1=3 \\ \Rightarrow a_n&=n^2+n+1\end{align} Taking summation of the both sides \begi... dfrac{3^n-3}{2}=\dfrac{1}{2} 3^n-\dfrac{3}{2} \\ \Rightarrow a_n&=\dfrac{1}{2} 3^n-\dfrac{3}{2}+3 \because a_1
- Question 2 & 3 Exercise 5.1
- frac{99(100)}{2} \\ & =(33.50 .199)+(99.50) \\ & \Rightarrow 1.2+2.3+3.4+\ldots+99.100 \\ & =3368050 . \end{al... es as: $$ \begin{aligned} & a_j=99=1+2(j-1) \\ & \Rightarrow 2 j-1=99 \\ & \Rightarrow j=\frac{100}{2}=50 . \end{aligned} $$ The sum of the 50 terms of the series $$ \beg... }{6}-\frac{4(49)(50)}{2}+50 \\ & =161700-4900+50 \Rightarrow 1^2+3^2+5^2+ \\ & \ldots+99^2=156850 . \end{align
- Question 2 & 3 Exercise 5.2
- x+16 x^2+24 x^3+\ldots \quad \text { (3) } \\ & \Rightarrow x(1-x) S_{\infty}=x+8 x^2+16 x^3+24 x^4+\ldots(4)... ty}=1+(8-1) x+(16-8) x^2+(24-16) x^3+\ldots \\ & \Rightarrow(1-2 x+x^2) S_{\infty}=1+7 x+ 8(x^2+x^3+x^4+\ldots) \\ & \Rightarrow(x-1)^2 S_{\infty}=1+7 x+8\dfrac{x^2}{1-x} \\ & \Rightarrow S_{\infty}=\dfrac{1+7 x}{(x-1)^2}-\dfrac{8x^2}{(x-
- Question 1 Exercise 5.3
- 1}{2}[18+6 n-12] \\ & =\dfrac{n-1}{2}[6+6 n] \\ \Rightarrow a_n&=3(n^2-1)+a_1 \\ \Rightarrow a_n&=3 n^2-3+4 \because a_1=4 \\ \Rightarrow a_n&=3 n^2+1 \end{align} Taking summation of the both sides \be... +1] \\ & =n \cdot[\dfrac{2 n^2+3 n+1+2}{2}] \\ & \Rightarrow \sum_{r=1}^n a_r=\dfrac{n}{2}(2 n^2+3 n+3)\\ \tex
- Question 3 Exercise 5.3
- 2}[2 n+8]=2 \cdot \dfrac{n-1}{2} \cdot[n+4] \\ & \Rightarrow a_n=n^2+3 n-4+a_1 \\ & \Rightarrow a_n=n^2+3 n-4+4 \quad \because a_1=4 \\ & \Rightarrow a_n=n^2+3 n\end{align} Taking summation of the both sid... dfrac{2 n+10}{3}] \\ & =\dfrac{n}{3}(n+1)(n+5)\\ \Rightarrow \sum_{r=1}^n a_r&=\dfrac{n}{3}(n+1)(n+5)\\ \text{
- Question 4 Exercise 5.3
- erms } \\ & =\dfrac{2 \cdot[3^{n- 1}-1]}{3-1} \\ \Rightarrow a_n-a_1&=3^{n-1}-1 \\ \Rightarrow a_n&=3^{n-1}-1+a_{1} \\ & \Rightarrow a_n=3^{n-1}-1+3=3^{n-1}+2 \end{align} Taking summation of the both side... }^n 1 \\ & =\dfrac{1 \cdot[3^n-1]}{3-1}+2 n \\ & \Rightarrow \sum_{r=1}^n a_r=\dfrac{1}{2}(3^n-1)+2 n \\ \text
- Question 6 Exercise 5.3
- xt { terms } \\ & =\dfrac{4[5^{n -1}-1]}{5-1} \\ \Rightarrow a_n \quad a_1&=5^{n-1}-1 \\ \Rightarrow a_n&=5^{n-1}-1+28 \quad \because a_1=28 \\ \Rightarrow a_n&=5^{n-1}+27\end{align} Taking summation of the bot... 1}^n 1 \\ & =\dfrac{1 \cdot[5^n-1]}{5-1}+27 n \\ \Rightarrow \sum_{r=1}^n a_r&=\dfrac{(5^n-1)}{4}+27 n\\ \text
- Question 1 Exercise 5.3
- frac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} \\ & \Rightarrow T_n=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking... }] \\ & =\dfrac{1}{2}(\dfrac{2 n+1-1}{2 n+1}) \\ \Rightarrow S_n&=\dfrac{n}{2 n+1}\end{align} =====Question 1... )$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Rightarrow(3 A+3 B) n+2 A-B&=1\end{align} Comparing the coef... )$ we get \begin{align} 1&=A(9 n+4)+B(9 n-5) \\ \Rightarrow(9 A+9 B) n+4 A-5 B&=1\end{align} Comparing the co
- Question 5 & 6 Review Exercise
- \ & =5+7 x+7 x^2+\cdots+7 x^{n-1}-(7 n-1) x^n \\ \Rightarrow(1-x) S_n&=5+7[x+x^2+\cdots+ x^{n-1}]-(7 n-1) x^n \\ \Rightarrow(1-x) S_n&=5+7 \cdot \dfrac{x(1-x^{n-1})}{1-x}-(7 n-1) x^n \\ \Rightarrow S_n&=\dfrac{5}{1-x}+\dfrac{7(x-x^n)}{(1-x)^2}-\df
- Question 10 Review Exercise
- frac{1}{2}} \\ a_n&=2 \cdot[1-\dfrac{1}{2^n}] \\ \Rightarrow a_n&=2-\dfrac{1}{2^{n-1}}\\ \Rightarrow a_n&=2(1-\dfrac{1}{2^{n}})\end{align} Taking summation on the both si
- Question 1 Exercise 5.1
- the both sides of the above, we get \begin{align}\Rightarrow \sum_{j=1}^n T_j&=\dfrac{1}{6} \sum_{j+1}^{j=n} j
- Question 9 Exercise 5.1
- \begin{align} & T_n=n^2(2 n+3)=2 n^3+3 n^2 \\ & \Rightarrow T_j=2 j^3+3 j^2\end{align} Taking sum of the both