# Thread: BASIC: How to Simpliy Powers for Induction

1. ## 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.

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

3. 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?

4. 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.

5. 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.

6. Why don't you show what you have done, then I will be able to help you further.

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

8. Nevermind, I managed to expand it and it managed to work out. Thanks for the help.

9. 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.