Using the Induction method, prove for n-1

1 + 3 + 9 + 27+ . . . 3n-1= (3n-1)/2

1*20 + 2*21 + 3*22 + . . . + n*2n-1 = (n-1)2n + 1

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- Sep 14th 2010, 07:37 AMtammyrProve Identities with Induction method
*Using the Induction method, prove for n-1*

1 + 3 + 9 + 27+ . . . 3n-1= (3n-1)/2

1*20 + 2*21 + 3*22 + . . . + n*2n-1 = (n-1)2n + 1

- Sep 14th 2010, 08:29 AMTraveller
You copied the sums wrong.

Take a look at the first few terms of each series and see whether they match the general formula that you provide at the end of each LHS ? Once you realize what the formulae are, prove them for n=1. Then assume that they are true for n-1 and add the general term for n to both sides of the equations. You should be able to get the rest of it by yourself. - Sep 14th 2010, 10:08 AMSoroban
Hello, tammyr!

C'mon . . . give us a break!

Are we supposed to GUESS what you meant?

Quote:

Using the Induction method, prove for n-1 ?

. . 1 + 3 + 9 + 27+ . . . 3n-1 = (3n-1)/2

. . 1*20 + 2*21 + 3*22 + . . . + n*2n-1 = (n-1)2n + 1

I'll take a guess at what you intended . . .

$\displaystyle 1 + 3 + 9 + 27 + \hdots + 3^{n-1} \;=\;\dfrac{3^n-1}{2}$

$\displaystyle 1\cdot2^0 + 2\cdot2^1 + 3\cdot2^2 + \hdots + n\cdot2^{n-1} \;=\;(n-1)2^n + 1$