Question

The population of a local species of beetle can be found using an infinite geometric series where a1 = 880 and the common ratio is one forth. Write the sum in sigma notation and calculate the sum (if possible) that will be the upper limit of this population.

Answer 1

@mathmale

Answer 2

I see immediately that the first term, a, is 880 and that the common ratio, r, is 1/4.

Can we use the geometric series sum formula? Yes, because the common ratio, 1/4, is between -1 and 1; in other words, the absolute value of the common ratio is |1/4|<1. Great.

Therefore, the sum of this series is

\[\sum_{n=1}^{infinity} a*r ^{n-1}\]

Answer 3

Please substitute 880 for a and 1/4 for r. Calculate this sum.

Answer 4

220

Answer 5

is this series divergent

Answer 6

\[\sum_{n=1}^{infinity} 880(\frac{ 1 }{ 4 }) ^{n-1}=?\]

Answer 7

Have you a book or online reference handy? We h ave not discussed this topic before. If the condition |r|<1 is met, then the sum in question is equal to \[\frac{ a }{ 1-r }\]

Answer 8

Again, a=880 and r=1/4. show your stuff!

Answer 9

880*1/4^0=880

Answer 10

Have you a book or online reference handy? We h ave not discussed this topic before. If the condition |r|<1 is met, then the sum in question is equal to
\[\frac{ a }{ 1-r }\]

Answer 11

880 is just the first term. a = 880.
1/4 is the common ratio. r=(1/4).

thus, substituting these values into

\[\frac{ a }{ 1-r }\]

Answer 12

what do you get? If done correctly, this result will represent the sum of the series.

Answer 13

Again, if |r|<1, then\[\sum_{n=1}^{infinity}ar ^{n-1}=\frac{ a }{ 1-r }\]

Answer 14

880/1-1/4

Answer 15

1173!!!

Answer 16

Would you please write this down. You'll defnitely see it again later.

You're on the right track, but what you have there is open to misinterpretation. Please use the Draw utility: