Question

The total number of fungal spores can be found using an infinite geometric series where a1 = 8 and the common ratio is 4. Find the sum of this infinite series that will be the upper limit of the fungal spores

Answer 1

Answer:

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$

Step-by-step explanation:

Explanation:-

Geometric sequence:-

The geometric sequence is of the form

$a,ar,ar^{2},ar^{3}$ ……… be an infinite sequence

here first term is 'a' and ratio is 'r '

In this geometric sequence 'n' t h term is $t_{n} = a r^{n-1}$....(1)

Given data  a1 term is '8 ' and ratio is '4'

substitute n=1 in equation(1)

$t_{1} = a r^{1-1}$ = 8

$ar=8$...........(2)

substitute r= 4 in equation(2)

now we get        a(4)=8

dividing "4" on both sides , we get   a = 2

Geometric series:-

$a+ar+ar^{2}+ar^{3}+$.....is an infinite geometric series.

sum of this infinite series will be the upper limit of the fungal spores

that is we have to find sum of infinite series

$S_{n} = \frac{a(r^{n}-1) }{r-1} if r>1$

$S_{n} = \frac{2(4^{n}-1) }{4-1} if r>1$

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$

Final answer:-

sum of this infinite series will be the upper limit of the fungal spores

$S_{n} = \frac{2(4^{n}-1) }3} if r>1$