Induction (1 Viewer)

Lukybear · Jan 29, 2010

Prove using induction

$-4 + 8 - 16 + ... + -4(-2)^ {n-1} = \frac{4((-2)^{n} - 1)}{3}$

The original document said 4(-2)^(n-1). And that works perfectly for induction. However, a = -4, and thus Tn = -4(-2)^(n-1). Please help.

EDIT: Fatal Error: 4((-2)^(n) - 1)/3 = RHS

Lukybear · Jan 29, 2010

yep ur absolutely rite. Now how do you do it for n?

And to check, it is on the LHS, it is -4 rite?

Lukybear · Jan 30, 2010

bobt2062 said:
Just showing the inductive step:

$S_{n+1}=S_n+T_{n+1}$
$=\frac{4(-2)^n-1}{3}-4(-2)^{n-1}$
$=\frac{4(-2)^n-1-12(-2)^{n-1}}{3}$
$=\frac{-8(-2)^n-1}{3}$
$=\frac{4(-2)^{n+1}-1}{3}$

Are you sure this is rite? I think youve done
$T_{n+1}$ wrong

It should be
$-4(-2)^{k+1)[tex] Also, the whole thing is in brackets. Sorry about mistake. Re-did it in edit.$

Induction (1 Viewer)

Lukybear

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Lukybear

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Lukybear

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