I have to find a little riddle , but i don't find a résulat.
can you help me please ?!
thx you!
It does make sense, actually; it is simply a hint that tells us the result is constant:
1. $\displaystyle (8^{n+1} + 8^n)^2 = ((2^3)^{n+1} + (2^3)^n)^2 = (2^{3n+3} + 2^{3n})^2 = (2^{3n}(1 + 8))^2 = \boxed{2^{6n}\cdot 81}$
2. $\displaystyle (4^{n}-4^{n-1})^3 = ((2^2)^n - (2^2)^{n-1})^3 = (2^{2n}-2^{2n-2})^3 = (2^{2n}(1 - \frac{1}{4}))^3 = \boxed{2^{6n}(\frac{27}{64})}$
Therefore $\displaystyle A = \frac{2^{6n}\cdot 81}{2^{6n}\cdot \frac{27}{64}} = 81 \cdot \frac{64}{27} = 3 \cdot 64 = 192$
Therefore $\displaystyle \boxed{A \equiv 192}$
You can plug in any value of $\displaystyle n\in \mathbb{N}, n >1$ and you will see that this is what you'll get.
Yes you've got to simplify the expression and you will find all the n's evenyuallu cancel out. I got 216 as the value for A.
Start by writing 8 as 2^3 and 4 as 2^2.
Then use your index laws and factorising.
A few hints: (2^3)^(n+1) = 2^(3n+3) = 2^(3n) x 2^3.
Give it a try....I'll check in again later if you need more help. Good luck.