# Need proof by induction

• September 8th 2013, 01:41 PM
Brent1986
Need proof by induction
(original question) Prove that the Summation of j=0 to n of (-1/2)^j = (2n+1 + (-1)n) / (3*2n)

I have a question that includes the following
P(k) (-1/2)0 + (-1/2)1 + (-1/2)2 + ... + (-1/2)k = (2k+1 + (-1)k) / (3*2k)

So I have k+1 as

P(k+1) ((2K+1 + (-1)k) / (3*2k)) + (-1/2)k+1 = (2k+2 +(-1)k+1) / (3*2k+1)
• September 8th 2013, 01:58 PM
Plato
Re: Need proof by induction
Quote:

Originally Posted by Brent1986
(original question) Prove that the Summation of j=0 to n of (-1/2)^j = (2n+1 + (-1)n) / (3*2n)

I have a question that includes the following
P(k) (-1/2)0 + (-1/2)1 + (-1/2)2 + ... + (-1/2)k = (2k+1 + (-1)k) / (3*2k)

So I have k+1 as

P(k+1) ((2K+1 + (-1)k) / (3*2k)) + (-1/2)k+1 = (2k+2 +(-1)k+1) / (3*2k+1)

$\\\sum\limits_{j = 0}^{N + 1} {{{\left( {\frac{{ - 1}}{2}} \right)}^j}} = \\\sum\limits_{j = 0}^N {{{\left( {\frac{{ - 1}}{2}} \right)}^j}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 2}} + 2{{( - 1)}^N} + {{3( - 1)}^{N+1}}}}{{3 \cdot {2^{N + 1}}}}\\=??$
• September 8th 2013, 02:40 PM
Brent1986
Re: Need proof by induction
Im sorry, im feeling kinda dumb can you tell me how I add the 2 equations you have on that 3rd line?
• September 8th 2013, 03:00 PM
Plato
Re: Need proof by induction
Quote:

Originally Posted by Brent1986
Im sorry, im feeling kinda dumb can you tell me how I add the 2 equations you have on that 3rd line?

Notice that $\frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} = \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + \frac{{{{( - 1)}^{N + 1}}}}{{{2^{N + 1}}}}$

So the $\text{LCD is }3\cdot 2^{N+1}$ thus $\frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + \frac{{{{( - 1)}^{N + 1}}}}{{{2^{N + 1}}}} = \frac{{{2^{N + 2}} + 2{{( - 1)}^N} + 3{{( - 1)}^{N + 1}}}}{{3 \cdot {2^{N + 1}}}}$

Now notice that $2{( - 1)^N} + 3{( - 1)^{N + 1}} = {( - 1)^{N + 1}}\left[ { - 2 + 3} \right]$.

BTW: I do not usually just hand out solutions.
• September 8th 2013, 03:18 PM
Brent1986
Re: Need proof by induction
That last part looks similar to the distributive property but our 'a' isn't the same (N vs. N +1) and the b is inverted. Is there a name for that property? Or is it just something I should remember.

I really appreciate the help, not looking for solutions just understanding!
• September 8th 2013, 03:40 PM
Plato
Re: Need proof by induction
Quote:

Originally Posted by Brent1986
That last part looks similar to the distributive property but our 'a' isn't the same (N vs. N +1) and the b is inverted. Is there a name for that property? Or is it just something I should remember.

There are several ways to do this:
$2{( - 1)^N} + 3{( - 1)^{N + 1}} = {( - 1)^N}\left[ {2 - 3} \right] = {( - 1)^N}\left[ { - 1} \right] = {( - 1)^{N + 1}}$
• September 8th 2013, 03:51 PM
Brent1986
Re: Need proof by induction
Much thanks to you my friend! I spent a few hours today trying to wrap my brain around these questions. I feel like I have a much better understanding now! You've really helped me!