1) Common ratio = \(\dfrac{6r}{2}\) = 3r
-1 < (Common ratio) < 1
-1 < 3r < 1
\(\dfrac{-1}{3} < r < \dfrac{1}{3}\)
Please explain. I do not understand what you have just said.
\(\text{Let }a\text{ be the first term.}\\ \dfrac{a}{1 - \dfrac{-1}{5}}= 16\\ a = 16\left(\dfrac{6}{5}\right) = \dfrac{96}{5}\)
The first term of the series is 96/5
Distribute one candy to each kids first.
5 candies are left, and I can distribute to them in any configuration.
I have 55 = 3125 ways to do so.
\(\quad (3 + 5x)^7\\ = \displaystyle \sum^7_{k = 0}\binom7k \cdot 3^{7 - k} \cdot (5x)^k\\ = \displaystyle \sum^7_{k = 0} \binom7k \cdot 3^7 \cdot \left(\dfrac{5}{3}\right)^k\cdot x^k\\ \text{When }k = 2,\\ \quad\text{Term}\\ = \displaystyle\binom72 \cdot 3^7 \cdot \left(\dfrac{5}{3}\right)^2 \cdot x^2\\ = 21 \cdot 243 \cdot 25 x^2\\ \quad\text{Coefficient of }x^2\\ = 21\cdot 243\cdot 25\\ =127575\)
\(a\mathbf{v} + b\mathbf{w} + c\mathbf{x} = \vec{0}\\ \begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&2&8\\0&4&16 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \text{Let }c = k\\ b +4k = 0\\ \boxed{b = -4k}\\ a - 4k - k = 0\\ \boxed{a = 5k}\\ \dfrac{a - b}c = \dfrac{5k-(-4k)}{k} = 9\)
\(\begin{array}{rl} &k^2 - 81\\ =&k^2 \color{red} - 9k + 9k\color{black} -81\\ =&(k^2 - 9k) + (9k - 81)\\ =&\color{red}k\color{black}(k-9)\color{red}+9\color{black}(k-9)\\ =&(k - 9)(\color{red}k + 9\color{black})\\ =&(k -9)(k+9) \end{array}\)
I was careless. I will attempt that again.
(WRONG ANSWER: DELETED)
Denote \(r_n\) the nth repunit.
\(\begin{array}{rcl} S &=& \displaystyle \sum^{100}_{n = 1} r_n\\ &=& \displaystyle \sum^{100}_{n = 1} \sum^{n-1}_{k = 0} 10^{k}\\ &=& \displaystyle \frac{1}{10}\sum_{n = 1}^{100} \sum_{k = 1}^{n} 10^{k}\\ &=& \displaystyle \dfrac{1}{10} \sum_{1 \leq k \leq n \leq 100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \sum_{n=k}^{100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \left(10^{k} (101 - k)\right)\\ &=& \displaystyle \dfrac{1}{10}\left(101\left(\dfrac{10\left(10^{100} - 1\right)}{10 - 1}\right)-\sum_{k= 1}^{100}k\cdot 10^{k}\right) \end{array}\\ \text{Let }S_{AG} = \displaystyle \sum_{k = 1}^{100} k\cdot 10^{k}\\ \begin{array}{rcl} S &=& \dfrac{101}{9}\left(10^{100} - 1\right) - \dfrac{1}{10}S_{AG}\\ S_{AG} &=& 1\cdot 10^1 + 2\cdot 10^2 + \cdots + 100\cdot 10^{100}---(1)\\ \dfrac{1}{10}S_{AG} &=& 1 + 2\cdot 10^1+\cdots +100\cdot 10^{99}---(2)\\ (1)-(2):\dfrac{9}{10}S_{AG} &=& 100\cdot 10^{100} - (1 + 10^1 + 10^2 +\cdots + 10^{99})\\ &=&10^{102} - \dfrac{10^{100} - 1}{9}\\ S_{AG} &=& \dfrac{10^{103}}{9} - \dfrac{10^{101} - 10}{81}\\ S &=& \dfrac{101}{9} (10^{100} - 1) - \left(\dfrac{10^{102}}{9}-\dfrac{10^{100} - 1}{81}\right)\\ &=& \dfrac{10^{100}}{9} - \dfrac{101}{9}+\dfrac{10^{100} - 1}{81}\\ &=& \dfrac{10^{100} - 1}{9}+\dfrac{10^{100} - 901}{81} \end{array}\\\)
