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Compute the sum1 + \frac{1}{3} + \frac{2}{3} + \frac{2}{9} + \frac{4}{9} + \frac{4}{27} + \frac{8}{27} + \frac{8}{81} + \dotsb

 Jan 25, 2015

Best Answer 

 #7
avatar+21842 
+5

P.S.

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\
= 1+ \frac{1}{3} + \frac{2}{3}
+ \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}
+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}
+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}
+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}
+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}
+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}
+\dots \ \\\\\\
= 1+ 1
+ \frac{2}{3}}
+ \frac{2^2}{3^2}}
+ \frac{2^3}{3^3}}
+\dots \ \\\\\\
s=
1+1*( \frac{2}{3} ) ^0
+1*(\frac{2}{3})^1
+1*(\frac{2}{3})^2
+1*(\frac{2}{3})^3
+1*(\frac{2}{3})^4+
\dots \$$

.
 Jan 26, 2015
 #1
avatar+21842 
+5

$$\small{\text{
sum
$
s=
1+1*( \frac{2}{3} ) ^0+
+1*(\frac{2}{3})^1+
+1*(\frac{2}{3})^2+
+1*(\frac{2}{3})^3+
+1*(\frac{2}{3})^4+
\dots
$
}}\\
a= 1 \\
r = \frac{2}{3} \\
s = 1 + \frac{a}{1-r} = 1 + \frac{1}{1-\frac{2}{3}} = 1 + \frac{1} {\frac{1}{3}} = 1 + 3 = 4 \\
\boxed{s= 4}$$

 

 Jan 25, 2015
 #2
avatar+99272 
+5

There are 2 GPs here

 

$$\\\[ 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{81}+\dotsb \]\\\\
S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3\\\\$$

-----------------------------------------

 

$$\\\[ \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}+\dotsb \]\\\\
S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{2}{3}}
=\frac{\frac{1}{3}}{\frac{1}{3}}=1\\\\\\
Total=3+1=4$$

.
 Jan 25, 2015
 #3
avatar+99272 
0

Again, we are both correct but heureka's method is preferable.        Mine is pretty silly really   

Thanks Heureka      

 Jan 25, 2015
 #4
avatar+98091 
+5

Here's another possibility...

Notice that the 2nd and 3rd terms sum to 1

And the 4th and 5th terms sum to 2/3

And the 6th and 7th terms sum to 4/9

So we have....adding in the first term.....

1 + 1/[1-(2/3)] = 1 + 1/(1/3)  = 1 + 3  = 4

 

 Jan 25, 2015
 #5
avatar+99272 
+5

There you go "All roads lead to Rome"    

 Jan 26, 2015
 #6
avatar+98091 
0

True, Melody........but not all zeroes do.....!!!

(The Troll can attest to this...)

 

 Jan 26, 2015
 #7
avatar+21842 
+5
Best Answer

P.S.

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\
= 1+ \frac{1}{3} + \frac{2}{3}
+ \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}
+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}
+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}
+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}
+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}
+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}
+\dots \ \\\\\\
= 1+ 1
+ \frac{2}{3}}
+ \frac{2^2}{3^2}}
+ \frac{2^3}{3^3}}
+\dots \ \\\\\\
s=
1+1*( \frac{2}{3} ) ^0
+1*(\frac{2}{3})^1
+1*(\frac{2}{3})^2
+1*(\frac{2}{3})^3
+1*(\frac{2}{3})^4+
\dots \$$

heureka Jan 26, 2015
 #8
avatar+99272 
0

Very nice Heureka     

 Jan 26, 2015

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