Question

1-36) what is the sum of 2 + (1/5 + 1/5² + 1/5³ +...,.. +... 1/(5 to power n) ..)
help how

Answer 1

`\displaystyle (9)/(4)`

Explanation:

`\displaystyle 2 + ((1)/(5^1) + (1)/(5^2) + (1)/(5^3) + ... + (1)/(5^n) )`

The second term of this sum is an Infinite Geometric Progression
To find the sum of an Infinite Geometric Progression,

- Find the first term ( 1/5 ) of the progression and call it a
- Find the common ratio, (divide any term by it's predecessor. i.e
`\displaystyle ((1)/(5^2) )/((1)/(5) ) = (5)/(5^2) = (1)/(5)`) and call it r

Finding the sum of given Geometric Progression:
Sum of Infinite GP:
`\displaystyle (a)/(1-r)`

Plugging our values in this equation, we get:
Sum of Geometric Progression =
`\displaystyle ((1)/(5) )/(1 - (1)/(5) ) = ((1)/(5) )/((5 - 1)/(5) ) = (1)/(4)`

Adding to find the actual answer:

`\displaystyle ((1)/(5^1) + (1)/(5^2) + (1)/(5^3) + ... + (1)/(5^n) ) = (1)/(4)`

`\displaystyle 2 + ((1)/(5^1) + (1)/(5^2) + (1)/(5^3) + ... + (1)/(5^n) ) = 2 + (1)/(4)`

`\displaystyle 2 + ((1)/(5^1) + (1)/(5^2) + (1)/(5^3) + ... + (1)/(5^n) ) = (9)/(4)`