Search
You can find the results of your search below.
Fulltext results:
- Question 7 and 8, Exercise 4.8
- dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = ... rac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*} Using value of $A$ a... ign*} This gives \begin{align*} T_k &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*}
- Question 11 and 12, Exercise 4.8
- {align*} Thus, \begin{align*} T_k &= \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*... n &= \sum_{k=1}^n T_k = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*... ping series, so most terms cancel out, and we are left with: \begin{align*} S_n &= \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)
- Question 3 and 4, Exercise 4.8
- _{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n} \\ & -\left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right) \end{align... terms }) \\ & =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & ... um_{k=1}^{n} \frac{3^{k}-1}{2} \\ & =\frac{1}{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
- Question 7 and 8, Exercise 4.1
- m, $a_{10}$ and the 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)^{
- Question 7 and 8, Exercise 4.5
- rms of a geometric 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}
- Question 14, 15 and 16, Exercise 4.7
- sum_{k=1}^{n} k \\ & = \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 ser
- Question 21 and 22, Exercise 4.7
- 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \\ & = 3\left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2... -2 \sum_{k=1}^{n} k -3 \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
- Question 5, 6 and 7, Exercise 4.4
- {n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(27)\left(-\frac{1}{3}\right) = -9 \\ & a_{3}=a_{1} r^{2}=(27)\left(-\frac{1}{3}\right)^{2} = 27 \cdot \frac{1}{9} = 3 \\ & a_{4}=a_{1} r^{3}=(27)\left(-\frac{1}{3}\right)^{3} = 27 \cdot \left(-\frac{1}{27}\right) = -1 \end{align*} Hence $a_1=27$, $a_2=-
- Question 9 and 10, Exercise 4.5
- eq 1.$$ Thus \begin{align*} S_4 & =\frac{343-(-1)\left(-\frac{1}{7}\right)}{1+\frac{1}{7}} \\ &=\frac{\f... }\\ \implies & r^3=\frac{1}{8} \\ \implies & r^3=\left(\frac{1}{2} \right)^3\\ \implies & r=\frac{1}{2}.... a_3 = a_1 r^{2}\\ \implies &\frac{3}{4} = a_1 \left(\frac{1}{2}\right)^{2} \\ \implies &\frac{3}{4} =... terms of a geometric series is $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \
- Question 17 and 18, Exercise 4.7
- } -6 \sum_{k=1}^{n} k + \sum_{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}\lef
- Question 3 and 4, Exercise 4.5
- $n$ terms of 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} \\ &= \... rms of a geometric series is\\ $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \
- Question 5 and 6, Exercise 4.5
- rms of a geometric 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} ... terms of a geometric series is $$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$ Thus, \
- Question 25 and 26, Exercise 4.7
- rac{1}{7}} + 3 \cdot \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... rac{1}{2}} + 6 \cdot \frac{1}{2} \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\left(1 - \frac{1}{2}\right
- Question 9 and 10, Exercise 4.8
- +1)(k+2)} = \sum_{k=3}^n T_k \\ &= \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
- Question 15 and 16, Exercise 4.1
- Find the indicated 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 \\