Question

The population of a local species of mosquitos can be found using an infinite geometric series where a1 = 740 and the common ratio is one sixth. Write the sum in sigma notation and calculate the sum (if possible) that will be the upper limit of this population.

Answer 1

@Hero

Answer 2

the summation of 740 times one sixth to the i minus 1 power, from i equals 1 to infinity. ; the sum is divergent
the summation of 740 times one sixth to the i minus 1 power, from i equals 1 to infinity. ; the sum is 888
the summation of 740 times one sixth to the i power, from i equals 1 to infinity. ; the series is divergent
the summation of 740 times one sixth to the i power, from i equals 1 to infinity. ; the sum is 88

Answer 3

@SolomonZelman

Answer 4

any idea what this means
a1 = 740 ?

Answer 5

Im guessing it goes in some kind of formula

Answer 6

a1 is short hand for the name of the first term.

an infinite geometric series
is
\[ a_1 + a_2+ a_3 + ...\]
which goes on forever.

they are telling you the first term is 740

Answer 7

the common ratio is one sixth

that means the ratio of two consecutive terms is 1/6. For example
\[ \frac{a_2}{a_1} = \frac{1}{6} \]
if we "solve" for a2 we get
\[ a_2 = \frac{1}{6} \cdot 740 \]

Answer 8

also, (because the ratio of two consecutive terms is 1/6 )
\[ \frac{a_3}{a_2} = \frac{1}{6} \]
if we solve for a3 we get
\[ a_3 = \frac{1}{6}\cdot a_2 = \frac{1}{6}\cdot \frac{1}{6} a_1 = 740 \left(\frac{1}{6}\right)^2 \]

Answer 9

if we use those numbers in
\[ a_1 + a_2 + a_3 + ...\]
we get
\[ 740 + 740 \cdot \frac{1}{6} + 740 \cdot \left(\frac{1}{6} \right)^2+... \]

Answer 10

Alright

Answer 11

So it would look like this \[\sum_{i=1}^{\infty}=740(\frac{ 1 }{ 6 })^{i}\]

Answer 12

@phi

Answer 13

the sigma (Greek capital S, short for sum) you posted has i=1 at the bottom
that means the first term has i=1. in other words,
\[ 740 \left(\frac{1}{6}\right)^1 \]
but you want
\[ 740 + 740 \cdot \frac{1}{6} + 740 \cdot \left(\frac{1}{6} \right)^2+... \\
740 \cdot \left(\frac{1}{6} \right)^0+ 740 \cdot \left(\frac{1}{6} \right)^1 + 740 \cdot \left(\frac{1}{6} \right)^2+... \]

notice that (1/6) to the zero power is 1. anything to the 0 exponent is 1
does that pattern help you figure out what the sigma notation looks like ?

Answer 14

See this for if this series converges:
https://en.wikipedia.org/wiki/Geometric_series#Common_ratio

Answer 15

\[\sum_{i=1}^{\infty}(\frac{ 1 }{ 6 })^{i=1}\]'the sum is divergent
\[\sum_{i=1}^{\infty}(\frac{ 1 }{ 6 })^{i=1}\];the sum is 88
\[\sum_{i=1}^{\infty}(\frac{ 1 }{ 6 })^{i}\];the series is divergent
\[\sum_{i=1}^{\infty}(\frac{ 1 }{ 6 })^{i=1} \];the sum is 88

Answer 16

These are my answers @phi

Answer 17

Which would be the correct one ? @Hero

Answer 18

@Jim766