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 = (2^{n}^{+1} + (-1)^{n}) / (3*2^{n})

I have a question that includes the following

P(k) (-1/2)^{0} + (-1/2)^{1} + (-1/2)^{2} + ... + (-1/2)^{k} = (2^{k}^{+1} + (-1)^{k}) / (3*2^{k})

So I have k+1 as

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

But I dont know how to get those sides to be equivalent. Can anyone please help?

$\displaystyle \\\sum\limits_{j = 0}^{N + 1} {{{\left( {\frac{{ - 1}}{2}} \right)}^j}} = \\\sum\limits_{j = 0}^N {{{\left( {\frac{{ - 1}}{2}} \right)}^j}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 1}} + {{( - 1)}^N}}}{{3 \cdot {2^N}}} + {\left( {\frac{{ - 1}}{2}} \right)^{N + 1}} =\\ \frac{{{2^{N + 2}} + 2{{( - 1)}^N} + {{3( - 1)}^{N+1}}}}{{3 \cdot {2^{N + 1}}}}\\=??$

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?

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

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

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

BTW: I do not usually just hand out solutions.

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!

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:

$\displaystyle 2{( - 1)^N} + 3{( - 1)^{N + 1}} = {( - 1)^N}\left[ {2 - 3} \right] = {( - 1)^N}\left[ { - 1} \right] = {( - 1)^{N + 1}}$

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!