Question

The population of a local species of flies can be found using an infinite geometric series where a1 = 940 and the common ratio is one fifth. Write the sum in sigma notation, and calculate the sum (if possible) that will be the upper limit of this population. the summation of 940 times one fifth to the i minus 1 power, from i equals 1 to infinity. ; the sum is 1,175 the summation of 940 times one fifth to the i minus 1 power, from i equals 1 to infinity. ; the sum is divergent the summation of 940 times one fifth to the i power, from i equals 1 to infinity. ; the sum is 1,175 the summation of 940 times one fifth to the i power, from i equals 1 to infinity. ; the series is divergent

Answer 1

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 $a_1=940,common\; ratio,r=\frac{1}{5}$

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

$\sum_{1}^{\infty}ar^{n-1}=\frac{a}{1-r}$because r<1

Where a= First term of GP

r=Common ratio

Substituting the values then we get

Population=$\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}=$

$\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}=\frac{940}{1-\frac{1}{5}}$ r <1

$\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}$

=$\frac{940}{\frac{5-1}{5}}$

=$\frac{940}{\frac{4}{5}}$

=$\frac{940\times 5}{4}$

=$235\times 5$

Population=$\sum_{1}^{\infty}940(\frac{1}{5})^{i-1}$=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.

Answer 2

In sigma notation, the sum is

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

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