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# help

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Find the value of k so that $$3 + \frac{3 + k}{4} + \frac{3 + 2k}{4^2} + \frac{3 + 3k}{4^3} + \dotsb = 8.$$

Feb 16, 2020

#1
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By arithmetico-geometric series,

$$a_1 g_1 + (a_1 + d)(g_1 r) + (a_1 + 2d)(g_1 r^2) + \dots = \frac{dg_2}{(1 - r)^2} + \frac{x_1}{1 - r}$$

(see https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series)

This gives an answer of k = 7.

Feb 16, 2020
#2
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Find the value of k so that
$$3 + \dfrac{3 + k}{4} + \dfrac{3 + 2k}{4^2} + \dfrac{3 + 3k}{4^3} + \dotsb = 8$$.

$$\begin{array}{|rcll|} \hline 3*1 + (3 + k)\left(1*\frac{1}{4}\right) + (3 + 2k)\left(1* \frac{1}{4^2} \right) + (3 + 3k)\left(1* \frac{1}{4^3} \right) + \dotsb + (3 + (n-1)k)\left(1*\left(\frac{1}{4}\right)^{n-1}\right) + \dotsb = 8 \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}3}+({\color{red}3}+k)+({\color{red}3}+2k)+({\color{red}3}+3k)+\dotsb \\ \boxed{a={\color{red}3},\ k \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}1}+{\color{blue}1}*{\color{green}\frac{1}{4}}+{\color{blue}1}*({\color{green}\frac{1}{4}})^2+{\color{blue}1}*({\color{green}\frac{1}{4}})^3+\dotsb \\ \boxed{ b={\color{blue}1},\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\ \hline \end{array}$$

The sum  of the first n terms of an arithmetico–geometric sequence has the form:

$$\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{ \dfrac{ ab-(a+nk)br^n} {1-r} + \dfrac{kbr(1-r^n)} {(1-r)^2}} \\ &&\boxed{a=3,\ b=1,\ r=\frac{1}{4}} \\\\ &=& \dfrac{3*1-(3+nk)*1*\frac{1}{4^n}}{1-\frac{1}{4}} + \frac{ k*1*\frac{1}{4}(1-\frac{1}{4^n})} { (1-\frac{1}{4})^2 } \\\\ &=&\frac{4}{3}\left( 3-\frac{3}{4^n}-\frac{nk}{4^n} \right) +\frac{16}{9}\left( \frac{k}{4}-\frac{k}{4*4^n} \right) \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{4*4^n}*\frac{4*4}{9} \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{9*4^{n-1}} \\\\ &=& 4+ \frac{4}{9}k -\frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right) \\ &&\boxed{\lim \limits_{n\to\infty} \frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right)=0} \\ 8 &=& 4+ \frac{4}{9}k \\ 4 &=& \frac{4}{9}k \quad | \quad : 4 \\ 1 &=& \frac{1}{9}k \quad | \quad * 9 \\ 9 &=& k \\ \mathbf{k} &=& \mathbf{9} \\ \hline \end{array}$$

Formula:

infinite arithmetico–geometric sequence

$$\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+k\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}$$

Feb 16, 2020
edited by heureka  Feb 17, 2020
edited by heureka  Feb 17, 2020