1is the answer = 1 ?
\hspace{-16}$\If $\mathbf{x=\left(8+3\sqrt{7}\right)^n\;,n\in\mathbb{N}}\;,$Then $\mathbf{x-x^2+x\left[x\right]=}$\\\\ Where $\mathbf{\left[x\right]=}$Greatest Integer Function
-
UP 0 DOWN 0 0 3
3 Answers
1actually i dunno how to solve it properly but,
the question can be re-written as (x)(1-{x}) ..where {x} is fractional part of x.
=> (8+3√7)n (1-{x})
and by some very randon intuition i guess, 1- {(8+3√7)n} = (8-3√7)n.
=> (8+3√7)n* (8-3√7)n
= 1..
waiting for someone to prove that 1- {(8+3√7)n} = (8-3√7)n ..
71If n is even integer then,
(8+3\sqrt{7})^{n}+(8-3\sqrt{7})^{n} = I+f+f'
where (8+3\sqrt{7})^{n}=I+f & (8-3\sqrt{7})^{n} = f' since (8-3\sqrt{7})^{n}<1
Since LHS is integer, RHS must be so, Hence f+f' = 1
Now Given expression as pointed out by rishabh is x(1-\left\{ x\right\})
which is (I+f)f'=(8+3\sqrt{7})^{n}(8-3\sqrt{7})^{n} =\boxed{1}
