First slide

Binomial theorem for negative integral and rational Index

Question

$If x is small so that x^{2} and higher powers can be neglected, then the$ $approximate value for \frac{(1 - 2 x)^{- 1} (1 - 3 x)^{- 2}}{(1 - 4 x)^{- 3}} is$

Moderate

A

1-2x
B

1-3x
C

1-4x
D

1-5x

Solution

$\begin{array}{l} \frac{(1 - 2 x)^{- 1} (1 - 3 x)^{- 2}}{(1 - 4 x)^{- 3}} \\ = (1 + 2 x) (1 + 6 x) (1 - 12 x) == (1 + 2 x + 6 x - 12 x) = 1 - 4 x \end{array}$

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