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The population of a local species of flies can be found using an infinite geometric series where \( a_1 = 940 \) and the common ratio is \(\frac{1}{5}\).

Write the sum in sigma notation, and calculate the sum (if possible) that will be the upper limit of this population.

A. \(\sum_{i=1}^{\infty} 940 \left(\frac{1}{5}\right)^{i-1}\); the sum is 1,175

B. \(\sum_{i=1}^{\infty} 940 \left(\frac{1}{5}\right)^{i-1}\); the sum is divergent

C. \(\sum_{i=1}^{\infty} 940 \left(\frac{1}{5}\right)^{i}\); the sum is 1,175

D. \(\sum_{i=1}^{\infty} 940 \left(\frac{1}{5}\right)^{i}\); the series is divergent

Answer :

Answer:

A.The summation of 940 times one fifth to the i minus 1 power ,from i equals to infinityb .'the sum is 1,175.

Step-by-step explanation:

We are given that the population of a local species of flies can be found using an infinite geometric series

Where [tex]a_1=940,common\; ratio,r=\frac{1}{5}[/tex]

We know that the formula of sum of infinite G.P

[tex]\sum_{1}^{\infty}ar^{n-1}=\frac{a}{1-r} [/tex]because r<1

Where a= First term of GP

r=Common ratio

Substituting the values then we get

Population=[tex]\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}=[/tex]

[tex]\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}=\frac{940}{1-\frac{1}{5}}[/tex] r <1

[tex]\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}[/tex]

=[tex]\frac{940}{\frac{5-1}{5}}[/tex]

=[tex]\frac{940}{\frac{4}{5}}[/tex]

=[tex]\frac{940\times 5}{4}[/tex]

=[tex]235\times 5[/tex]

Population=[tex]\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}[/tex]=1,175

Hence, the sum is 1,175

Hence, option A is true.

Answer : A.The summation of 940 times one fifth to the i minus 1 power ,from i equals to infinityb .'the sum is 1,175.

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Rewritten by : Barada

In sigma notation, the sum is

[tex]population=\displaysize\sum\limits_{i=1}^{\infty}{940\cdot\left(\dfrac{1}{5}\right)^{(i-1)}}=\dfrac{940}{1-\frac{1}{5}}=1175[/tex]

the summation of 940 times one fifth to the i minus 1 power, from i equals 1 to infinity; the sum is 1,175