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In the geometric series [infinity]Σn= 2^n/(-5)^n+1

we have r=___
Since |r|< _____
Then [infinity]Σn= 2^n/(-5)^n+1 converges and [infinity]Σn= 2^n/(-5)^n+1 _______/_____

User Greg Dan
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Answer: the common ratio (r) is -2/5, |r| is less than 1, and the sum of the given series Σn=2^n/(-5)^n+1 is -4/175.

Step-by-step explanation:

In the given geometric series Σn=2^n/(-5)^(n+1), we can determine the common ratio (r) and the condition for convergence by analyzing the ratio of consecutive terms.

The general form of a geometric series is Σn=0 to ∞ ar^n, where a is the first term and r is the common ratio.

Comparing the given series to the general form, we have:

a = 2^2/(-5)^3 = 4/(-125) = -4/125

To find the common ratio (r), we divide the (n+1)th term by the nth term:

r = (2^(n+1))/(-5)^(n+2) divided by 2^n/(-5)^(n+1)

= (2^(n+1))*((-5)^(n+1))/((-5)^(n+2))*2^n

= 2/(-5)

= -2/5

To ensure convergence, we need the absolute value of the common ratio (|r|) to be less than 1.

|r| = |-2/5| = 2/5 < 1

Since |r| is less than 1, the given series Σn=2^n/(-5)^n+1 converges.

To determine the sum of the series, we use the formula for the sum of an infinite geometric series:

Sum = a/(1 - r)

Plugging in the values, we have:

Sum = (-4/125)/(1 - (-2/5))

= (-4/125)/(1 + 2/5)

= (-4/125)/(5/5 + 2/5)

= (-4/125)/(7/5)

= (-4/125) * (5/7)

= -20/875

= -4/175

User Squiter
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