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

17 votes
17 votes

Answer:


\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)

User Kevin Suchlicki
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