# How do you use the binomial series to expand #(1 + 4x)^-3#?

From the formula for the binomial theorem

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The binomial series for ((1 + x)^n) where (|x| < 1) and (n) is any real number is given by: [ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots ]

To expand ((1 + 4x)^{-3}), we use (n = -3) and replace (x) with (4x): [ (1 + 4x)^{-3} = 1 + (-3)(4x) + \frac{(-3)(-4)}{2!}(4x)^2 + \frac{(-3)(-4)(-5)}{3!}(4x)^3 + \ldots ]

Simplifying each term: [ = 1 - 12x + \frac{12 \cdot 4}{2}(4x)^2 + \frac{-12 \cdot 20}{6}(4x)^3 + \ldots ] [ = 1 - 12x + 24(16x^2) - \frac{240}{6}(64x^3) + \ldots ] [ = 1 - 12x + 384x^2 - 2560x^3 + \ldots ]

Therefore, the series expansion of ((1 + 4x)^{-3}) is: [ 1 - 12x + 384x^2 - 2560x^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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