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How do you sum the series #1+a+a^2+a^3+...+a^n#?

Answer 1

Anna Allen

#1+a+a^2+a^3+...+a^n = (1-a^(n+1))/(1-a)#

#(1+a+a^2+a^3+...+a^n)(1-a)=1-a^(n+1)#

Divide both sides by #(1-a)# to get:

#1+a+a^2+a^3+...+a^n = (1-a^(n+1))/(1-a)#

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Answer 2

Christian Antunez

The solution George C.gives is very elegant. Here's another that has its own uses.

Let

#S_n = 1 + a + a^2 +a^3 + * * * +a^n#

Multiply by #a# to get:

#aS_n = " " a + a^2 +a^3 + * * * +a^n +a^(n+1)#

Now subtract:

#(1-a)S_n = 1-a^(n+1)#

So, again:

#S_n = (1-a^(n+1))/(1-a)#

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Answer 3

Greyson Brown

To sum the series (1 + a + a^2 + a^3 + \ldots + a^n), you can use the formula for the sum of a geometric series. The formula is:

[S = \frac{{a^{n+1} - 1}}{{a - 1}}]

Where:

(S) is the sum of the series.
(a) is the common ratio of the geometric series.
(n) is the number of terms in the series.

So, to find the sum of the given series, plug in the values of (a) and (n) into the formula and calculate accordingly.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.