Question Help Please (1 Viewer)

fake43 · Oct 28, 2021

Could someone help with (i) (2020 killara trial paper)

CM_Tutor · Oct 28, 2021

Sum the series as a GP

The resulting expression in t can then be integrated from t = 0 to t = x

Drongoski · Oct 28, 2021

If I recall correctly:

$\frac {1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum ^{ \infty} _ {i =0 } x^i $ for $ |x| < 1\\ \\ \frac {1}{1+x} = \frac {1}{1 - (-x)} = 1 - x + x^2 - x^3 + \cdots = \sum ^{\infty} _ {i=0} (-x)^i $ for $|x| < 1$

Question Help Please (1 Viewer)

fake43

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CM_Tutor

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Drongoski

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