+0

# Compute the sum

0
1814
8

Compute the sum

Jan 25, 2015

#7
+22358
+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 \$$

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Jan 26, 2015

#1
+22358
+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
+101769
+5

There are 2 GPs here

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

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

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

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Jan 25, 2015
#3
+101769
+1

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

Thanks Heureka

Jan 25, 2015
#4
+101424
+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
+101769
+5

Jan 26, 2015
#6
+101424
0

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

(The Troll can attest to this...)

Jan 26, 2015
#7
+22358
+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 \$$

heureka Jan 26, 2015
#8
+101769
0

Very nice Heureka

Jan 26, 2015