BASIC: How to Simpliy Powers for Induction

Printable View

May 30th 2011, 09:57 PM
stormerze

BASIC: How to Simpliy Powers for Induction

Hey guys, I'm just a bit unsure how to simplify numbers so that this induction can be proved. Any help would be appreciated.
$2\frac{{[k(k+1)}]^{ 4} }{ 16} + {(k+1)}^{5 }+ {(k+1)}^{7 }= 2\frac{{[(k+1)(k+2)]}^{4 } }{ 16}$

Thanks. Hope the latex came out alright.
May 30th 2011, 10:02 PM
Prove It

Start by getting a common denominator, then looking for common factors in the numerator.
May 30th 2011, 10:03 PM
stormerze

Quote:

Originally Posted by Prove It

Start by getting a common denominator, then looking for common factors in the numerator.

I'm only allowed to work with one side though. Are you suggesting I get everything divisible by 16?
May 30th 2011, 10:07 PM
Prove It

Quote:

Originally Posted by stormerze

I'm only allowed to work with one side though. Are you suggesting I get everything divisible by 16?

Exactly, on the LHS get a common denominator, then in the numerator, look for common factors. See if you can work on the LHS to get the stuff on the RHS.
May 30th 2011, 10:38 PM
stormerze

I'm mainly confused with simplifying with powers and thus don't know how to simplify the top after converting it with a base of 16.
May 30th 2011, 10:42 PM
Prove It

Why don't you show what you have done, then I will be able to help you further.
May 30th 2011, 11:12 PM
stormerze

Not 100% sure whether this is right or not but I simplified it to:
$\frac{{k}^{ 4}{(k+1)}^{ 4} +8[(k+1)+{(k+1)}^{ 3}]{(k+1)}^{4 } }{ 8}$
May 30th 2011, 11:22 PM
stormerze

Nevermind, I managed to expand it and it managed to work out. Thanks for the help. :)
May 30th 2011, 11:59 PM
Prove It

Quote:

Originally Posted by stormerze

Not 100% sure whether this is right or not but I simplified it to:
$\frac{{k}^{ 4}{(k+1)}^{ 4} +8[(k+1)+{(k+1)}^{ 3}]{(k+1)}^{4 } }{ 8}$

There's nothing wrong with expanding everything out and then factorising, but it is nearly always easier if you look for common factors first.

$\displaystyle \begin{align*}\frac{k^4(k + 1)^4 + 8[(k + 1) + (k + 1)^3](k + 1)^4}{8} &= \frac{(k+ 1)^4\{k^4 + 8[(k + 1) + (k + 1)^3]\}}{8} \\ &= \frac{(k + 1)^4\{k^4 + 8(k + 1) + 8(k + 1)^3\}}{8} \\ &= \frac{(k + 1)^4[k^4 + 8k + 8 + 8(k^3 + 3k^2 + 3k + 1)]}{8} \\ &= \frac{(k + 1)^4(k^4 + 8k + 8 + 8k^3 + 24k^2 + 24k + 8)}{8} \\ &= \frac{(k + 1)^4(k^4 + 8k^3 + 24k^2 + 32k + 16)}{8} \\ &= \frac{(k + 1)^4(1\cdot 2^0 \cdot k^4 + 4\cdot 2^1 \cdot k^3 + 6\cdot 2^2 \cdot k^2 + 4\cdot 2^3 \cdot k^1 + 1\cdot 2^4 \cdot k^0)}{8} \\ &= \frac{(k + 1)^4(k + 2)^4}{8}\end{align*}$

Alternatively you can apply the factor theorem to factorise the quartic.