# How do you find the Maclaurin series for #1/(2-3x)^2#?

converging for

Note that:

So that:

Now express:

we have:

In the interval of convergence we can differentiate the series term by term:

Or alternatively:

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To find the Maclaurin series for ( \frac{1}{{(2-3x)}^2} ), you can use the geometric series expansion formula:

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

First, rewrite the given function as:

[ \frac{1}{{(2-3x)}^2} = (2-3x)^{-2} ]

Now, substitute ( u = 3x - 2 ) into the geometric series formula:

[ \frac{1}{(2-3x)^2} = \frac{1}{u^2} ]

Using the geometric series formula:

[ \frac{1}{u^2} = 1 + u^2 + u^4 + u^6 + \dots ]

Substituting back for ( u ):

[ = 1 + (3x - 2)^2 + (3x - 2)^4 + (3x - 2)^6 + \dots ]

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