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- Question 7 and 8, Exercise 4.8
- e \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*} N... e \begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*} Using value of $A$ and $B$ in (1), we get \begi... &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*} Taking sum \begin{align*} S_n&=\sum
- Question 11 and 12, Exercise 4.8
- = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*} Taking the sum, we have \begin{ali... \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*} This is a telescoping series, so m... 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) \\ & = \end{align*} <fc #ff0000>This will be so... e \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*} N
- Question 3 and 4, Exercise 4.8
- {n} \\ & -\left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1... 1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =1+\frac{3^{n}-3}{2} \\ & =\frac{2+3^{n}-... }{2}\left(\sum_{k=1}^{n} 3^{k} - \sum_{k=1}^{n} 1\right) \\ & =\frac{1}{2}\left((3+3^2+3^3+\ldots +3^{n})-n\right) \\ & =\frac{1}{2}\left(\frac{3(3^n-1)}{3-1}-n\r
- Question 7 and 8, Exercise 4.1
- e 15 th term: $a_{15}$.$a_{n}=\left(\frac{-1}{2}\right)^{n-1}$ ** Solution. ** Given: $$a_n = \left( \frac{-1}{2} \right)^{n-1}.$$ Now \begin{align*}a_1 &= \left( \frac{-1}{2} \right)^{1-1} = \left( \frac{-1}{2} \right)^0 = 1 \\ a_2 &= \left( \frac{-1}{2} \right)^{2-1} = \left( \frac{-1
- Question 7 and 8, Exercise 4.5
- etric series is\\ $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \begin{align*} S_{10} &= \frac{16 \left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)} \\ &= \frac{16 \left(1 - \frac{1}{1024}\right)}{1 + \frac{1}{2}} \\ &=
- Question 14, 15 and 16, Exercise 4.7
- frac{n(n+1)(2n+1)}{6} + 2\left( \frac{n(n+1)}{2} \right) \\ & = \frac{n}{6}\left[(n+1)(2n+1)+6(n+1)\right] \\ & = \frac{n}{6}\left(2n^2+2n+n+1+6n+6\right) \\ & = \frac{n}{6}\left(2n^2+9n+7\right) \end{align*} Thus, the sum of the series is $\sum\limits_{k=1
- Question 21 and 22, Exercise 4.7
- k=1}^{n} k \\ & = 3\left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2} \\ & = \frac{n(n+1)}{2}(2n+1+... \sum_{k=1}^{n} 1 \\ & = 8\left(\frac{n(n+1)}{2}\right)^3+12\left(\frac{n(n+1)(2n+1)}{6}\right)-2\left( \frac{n(n+1)}{2} \right) - 3n \\ & = \left(n(n+1)\right)^3+2\left(n(n+1)(2n+1)\right)-\left( n(
- Question 5, 6 and 7, Exercise 4.4
- in{align*} & a_{2}=a_{1} r=(27)\left(-\frac{1}{3}\right) = -9 \\ & a_{3}=a_{1} r^{2}=(27)\left(-\frac{1}{3}\right)^{2} = 27 \cdot \frac{1}{9} = 3 \\ & a_{4}=a_{1} r^{3}=(27)\left(-\frac{1}{3}\right)^{3} = 27 \cdot \left(-\frac{1}{27}\right) = -1 \end{align*} Hence $a_1=27$, $a_2=-9$, $a_3=3$, $a_4=-1
- Question 9 and 10, Exercise 4.5
- n{align*} S_4 & =\frac{343-(-1)\left(-\frac{1}{7}\right)}{1+\frac{1}{7}} \\ &=\frac{\frac{2400}{7}}{\frac... =\frac{1}{8} \\ \implies & r^3=\left(\frac{1}{2} \right)^3\\ \implies & r=\frac{1}{2}. \end{align*} Now ... \\ \implies &\frac{3}{4} = a_1 \left(\frac{1}{2}\right)^{2} \\ \implies &\frac{3}{4} = a_1 \frac{1}{4}\\... ometric series is $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \begin{align*}
- Question 17 and 18, Exercise 4.7
- k=1}^{n} 1 \\ & = 9\left( \frac{n(n+1)(2n+1)}{6} \right) - 6\left( \frac{n(n+1)}{2} \right) + n \\ & = \frac{3n(n+1)(2n+1)}{2} -3 n(n+1) + n \\ & = \frac{n}{2}\left[3(n+1)(2n+1)-6(n+1) + 2\right]\\ & = \frac{n}{2}\left(6n^2+6n+3n+3-6n-6+2\right) \\ & = \frac{n}{2}\left( 6n^2+3n-1\right) \end{align
- Question 3 and 4, Exercise 4.5
- geometric series \[ S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r\neq 1. \] Thus \begin{align*} S_{12} &= \frac{5\left(1 - 3^{12}\right)}{1 - 3} \\ &= \frac{5\left(1 - 531441\right)}{-2} \\ &= \frac{5(-531440)}{-2} \\ &= \frac{-2657200}{-2}... etric series is\\ $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \begin{align*}
- Question 5 and 6, Exercise 4.5
- etric series is\\ $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \begin{align*} S_{14} &= \frac{7 \left(1 - 2^{14}\right)}{1 - 2} \\ &= \frac{7 \left(1 - 16384\right)}{-1} \\ &= \frac{7 \times (-16383)}{-1} \\ &= 7 \times 163... ometric series is $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \begin{align*}
- Question 25 and 26, Exercise 4.7
- dot \frac{1}{7} \cdot \frac{1 - \left(\frac{1}{7}\right)^n}{\left(1 - \frac{1}{7}\right)^2} - \frac{(1 + n \cdot 3) \left(\frac{1}{7}\right)^n}{1 - \frac{1}{7}}\\ &= \frac{7}{6} + \frac{21(1 -... dot \frac{1}{2} \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\left(1 - \frac{1}{2}\right)^2} - \frac{(1 +
- Question 9 and 10, Exercise 4.8
- sum_{k=3}^n \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \\ &= \left( \frac{1}{4} - \frac{1}{5} \right)+ \left( \frac{1}{5} - \frac{1}{6} \right)+\left( \frac{1}{6} - \frac{1}{7} \right) \\ &\quad +...+\left( \frac{1}{n} - \frac{1}{n+1} \right)+ \le
- Question 15 and 16, Exercise 4.1
- term of the sequence. $a_{n}=\left(1+\frac{1}{n}\right)^{2} ; a_{20}$ ** Solution. ** Given: $$a_n = \left(1 + \frac{1}{n}\right)^2.$$ Then \begin{align*} a_{20} &= \left(1 + \frac{1}{20}\right)^2 \\ &= \left(\frac{20 + 1}{20}\right)^2 \\ &= \left(\frac{21}{20}\right)^2 \\ &= \frac{441}{400} \end{