anonymous
  • anonymous
how to find the sum from 0 to infinity (2^(K+3))/3^K
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{0}^{\infty} 2^{k+3}/3^{k}\]
amistre64
  • amistre64
2^k 2^3/3^k
amistre64
  • amistre64
got it

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More answers

amistre64
  • amistre64
\[\sum2^3(\frac{2}{3})^k\to 8\sum (\frac{2}{3})^k \]
amistre64
  • amistre64
the sum part goes to 1/(1-2/3)
amistre64
  • amistre64
1/(1/3) = 3 8*3 = 24
anonymous
  • anonymous
great explanation! Got it man!
amistre64
  • amistre64
yw
anonymous
  • anonymous
what if the denominator goes from 3^k to 3^(K+3) and then numerator became 1
amistre64
  • amistre64
same concepts, just different numbers really
anonymous
  • anonymous
\[\sum_{0}^{\infty} 1/3^{k+3}\]
amistre64
  • amistre64
\[B^{a+b}\to B^aB^b\] that allow you to factor off the constant and use the rest as a ratio
amistre64
  • amistre64
1/3^3 sum (1/3)^k
amistre64
  • amistre64
the sum of a geometric sequence is:\[\sum \frac{1-r^n}{1-r}\to \frac{1-r^{inf}}{1-r}\] when r is less than 1, r^inf goes to zero
amistre64
  • amistre64
well, when |r|<1 ...
anonymous
  • anonymous
true now I'm struggling on this question \[\sum_{0}^{\infty} 3^{k-1}/4^{3k+1}\] i try to use your method, but didn't work I have no idea how to make the denominator good
amistre64
  • amistre64
well, lets start by stripping off the boring things :)
amistre64
  • amistre64
\[ \sum_{0}^{\infty} 3^{k-1}/4^{3k+1}\] \[ \sum_{0}^{\infty} 3^{k}3^{-1}/4^{3k}4^{1}\] \[ 1/12\sum_{0}^{\infty} 3^{k}/4^{3k} \] \[ 1/12\sum_{0}^{\infty} 3^{k}/(4^3)^k \] \[ 1/12\sum_{0}^{\infty} (3/64)^k \]
amistre64
  • amistre64
now r=3/64
anonymous
  • anonymous
oh ya!!

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