+0  
 
0
135
1
avatar

A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a$, $b$, and $c$ are integers. Find $a+b+c$.

Guest Sep 17, 2018
 #1
avatar+20591 
+8

A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a$, $b$, and $c$ are integers. Find $a+b+c$.

 

Formula:

\(\begin{array}{|rclcl|} \hline s_n &=& \dfrac{2}{3+1} + \dfrac{2^2}{3^2+1} + \dfrac{2^3}{3^{(2^2)}+1} + \dfrac{2^4}{3^{(2^3)}+1} + \dfrac{2^5}{3^{(2^4)}+1} + \ldots + \dfrac{2^{n+1}}{3^{(2^n)}+1} \\\\ s_0 &=& \dfrac12 \\ s_1 &=& \dfrac12 + \dfrac{4}{10} = \dfrac{9}{10} \\ s_2 &=& \dfrac{9}{10}+ \dfrac{8}{82}= \dfrac{409}{410} \\ s_3 &=& \dfrac{409}{410}+ \dfrac{16}{6562}= \dfrac{1345209}{1345210} \\ \ldots \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 - \dfrac{a^{n+b}}{3^{ (2^{n+c} )}-1} \\\\ s_n &=& 1 - \dfrac{a^n a^b} { 3^{ (2^n 2^c )}-1 } \\\\ s_n &=& 1 - \dfrac{a^n a^b} { \left(3^{(2^c)} \right)^{2^n}-1 } \\ && \text{Set }~ \alpha=a^b \\ && \text{Set }~ \gamma=3^{(2^c)} \\ \mathbf{s_n} & \mathbf{=} & \mathbf{1 - \dfrac{a^n \alpha} { \gamma^{(2^n)}-1 } } \\ \hline \end{array}\)

 

\(\begin{array}{rcll} \boxed{n=0}:\\ s_0 = \dfrac12 &=& 1 - \dfrac{a^0 \alpha} { \gamma^{(2^0)}-1 } \\ \dfrac12 &=& 1 - \dfrac{\alpha} { \gamma-1 } \\ \dfrac{\alpha} { \gamma-1 } &=& 1 - \dfrac12 \\ \dfrac{\alpha} { \gamma-1 } &=& \dfrac12 \\ \mathbf{ \alpha } & \mathbf{=}& \mathbf{ \dfrac12\cdot (\gamma-1)} \qquad (1) \\ \end{array} \)

 

\(\begin{array}{rcll} \boxed{n=1}:\\ s_1 = \dfrac{9}{10} &=& 1 - \dfrac{a^1 \alpha} { \gamma^{(2^1)}-1 } \\ \dfrac{9}{10} &=& 1 - \dfrac{a \alpha} { \gamma^2-1 } \\ \dfrac{a \alpha} { \gamma^2-1 } &=& 1 - \dfrac{9}{10} \\ a\alpha & = & \dfrac{1}{10}\cdot (\gamma^2-1) \quad & | \quad \gamma^2-1 = (\gamma-1)(\gamma+1) \\ \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \\ \end{array}\)

 

\(\begin{array}{rcll} \boxed{n=2}:\\ s_2 = \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^{(2^2)}-1 } \\ \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^4-1 } \\ \dfrac{a^2 \alpha} { \gamma^4-1 } &=& 1 - \dfrac{409}{410} \\ a^2 \alpha &=& \dfrac{1}{410}(\gamma^4-1) \quad | \quad \gamma^4-1 = (\gamma^2-1)(\gamma^2+1)= (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\ \end{array}\)

 

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \quad & | \quad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \qquad (1) \\ a\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1) \\ a\cdot \dfrac12 & = & \dfrac{1}{10}\cdot (\gamma+1) \\ \mathbf{a} & \mathbf{=} & \mathbf{\dfrac{1}{5}\cdot (\gamma+1)} \qquad (4) \\ \hline \end{array} \)


\(\mathbf{\gamma=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\\\ && \mathbf{a = \dfrac{1}{5}\cdot (\gamma+1)} \quad (4) \qquad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \quad (1)\\\\ \dfrac{1}{25}\cdot (\gamma+1)^2\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \dfrac{1}{25}\cdot (\gamma+1)\cdot \dfrac12 & = & \dfrac{1}{410}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\gamma^2+\dfrac{5}{41} \\ \dfrac{5}{41}\gamma^2-\gamma-1+\dfrac{5}{41} &=& 0 \\ \dfrac{5}{41}\gamma^2-\gamma- \dfrac{36}{41} &=& 0 \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1-4\cdot\dfrac{5}{41}\cdot \left(-\dfrac{36}{41} \right) } } {2\cdot \dfrac{5}{41} } \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1+ \dfrac{720}{41^2} } } {\dfrac{10}{41} } \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \sqrt{1+ \dfrac{720}{41^2} } \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ \sqrt{2401} } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ 49 } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{41}{10} \cdot \dfrac{49} {41} \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{49}{10} \\\\ \gamma &=& \dfrac{41+49}{10} \\\\ \gamma &=& \dfrac{90}{10} \\ \mathbf{ \gamma }& \mathbf{=}& \mathbf{9} \\\\ \gamma &=& \dfrac{41-49}{10} \\\\ \gamma &=& -\dfrac{8}{10} \qquad \text{no solution, } \gamma\gt 0\\ \hline \end{array} \)

 

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline a & = & \dfrac15\cdot (\gamma+1) \quad & | \quad \gamma = 9 \\ a & = & \dfrac15\cdot (9+1) \\ a & = & \dfrac{10}{5} \\ \mathbf{a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\mathbf{\alpha=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha & = & \dfrac12\cdot (\gamma-1) \quad & | \quad \gamma = 9 \\ \alpha & = & \dfrac12\cdot (9-1) \\ \alpha & = & \dfrac{8}{2} \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

 

\(\mathbf{b=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha &=& a^b \quad & | \quad a=2 \qquad \alpha = 4 \\ 4 &=& 2^b \\ 2^2 &=& 2^b \\ 2 &=& b \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\mathbf{c=~ ?} \)

\(\begin{array}{|rcll|} \hline \gamma &=& 3^{(2^c)} \quad & | \quad \gamma = 9 \\ 9 &=& 3^{(2^c)} \\ 3^2 &=& 3^{(2^c)} \\ 2 &=& 2^c \\ 2^1 &=& 2^c \\ 1 &=& c \\ \mathbf{c} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} && \mathbf{a+b+c} \\ &=& 2+2+1 \\ &=& \mathbf{5} \\ \end{array}\)

 

laugh

heureka  Sep 17, 2018

12 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.