Question

Find the sum of this geometric series

Answer 1

S₆ = 39060

Explanation:

the sum of a finite geometric sequence is

`(a_(1)(r^(n)-1) )/(r-1)`

where a₁ is the first term and r the common ratio

the expression inside the ∑ is not in this form

so we require to find a₁ and r

the common ratio r =
`(a_(n+1) )/(a_(n) )` , then

r =
`(2(5^(n+1)) )/(2(5^(n)) )` ← cancel 2 on numerator/ denominator

=
`(5^(n+1) )/(5^(n) )`

=
`(5^(n)(5) )/(5^(n) )` ← cancel
`5^(n)` on numerator/ denominator

= 5

find first term in series a₁ by substituting in the lower bound and simplifying.

a₁ = 2
`(5)^(1)` = 2 × 5 = 10

now substitute a₁ = 10 and r = 5 into the sum formula

`(10(5^(6)-1) )/(5-1)`

=
`(10(15625-1))/(4)`

=
`(10(15624))/(4)`

=
`(156240)/(4)`

= 39060