# How do you find the Maclaurin series for #1/(1+x)^3#?

I got

or

COMMON WAY

We can take the derivative more than once, and it is best to do this before writing out the sum in full.

Let us now examine what results: The Maclaurin series becomes:

OLDER METHOD; MORE TIME

Our actions are:

Step 1:

Step 2:

Step 3:

Step 4:

which Wolfram Alpha appears to concur with.

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To find the Maclaurin series for ( \frac{1}{(1+x)^3} ), we can use the geometric series expansion. The geometric series expansion is given by:

[ \frac{1}{1 - u} = 1 + u + u^2 + u^3 + \ldots ]

Substitute ( u = -x ) into the geometric series expansion:

[ \frac{1}{1 + x} = 1 - x + x^2 - x^3 + \ldots ]

Now, raise both sides to the power of 3:

[ \left(\frac{1}{1 + x}\right)^3 = (1 - x + x^2 - x^3 + \ldots)^3 ]

Expanding the right side using the binomial theorem:

[ = 1 - 3x + 6x^2 - 10x^3 + \ldots ]

Thus, the Maclaurin series for ( \frac{1}{(1+x)^3} ) is:

[ 1 - 3x + 6x^2 - 10x^3 + \ldots ]

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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.

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.

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.

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