Question

If $x={(\sqrt{3}+1)}^n$ such that $n\in N$ and is an odd number then $\left[x\right]$ is (where $\left[x\right]$ denotes the greatest integers less than or equal to $x$)

• A. $2k$, where $k\in I$
• B. $2k+1$, where $k\in I$
• C. $4^n$
• D. $8^n$

Answer 1

The correct option is A $2k$, where $k\in I$

$x=(\sqrt{3}+1{)}^n=$ ${}^n{C}_0\left(\sqrt{3}{)}^n\right(1{)}^0$ + ${}^n{C}_1\left(\sqrt{3}{)}^{n-1}\right(1{)}^1$ + ${}^n{C}_2\left(\sqrt{3}{)}^{n-2}\right(1{)}^2$ + ${}^n{C}_3\left(\sqrt{3}{)}^{n-3}\right(1{)}^3$ + ...... + ${}^n{C}_n\left(\sqrt{3}{)}^0\right(1{)}^n$

Let $y=(\sqrt{3}-1{)}^n$ = ${}^n{C}_0\left(\sqrt{3}{)}^n\right(-1{)}^0$ + ${}^n{C}_1\left(\sqrt{3}{)}^{n-1}\right(-1{)}^1$ + ${}^n{C}_2\left(\sqrt{3}{)}^{n-2}\right(-1{)}^2$ + ${}^n{C}_3\left(\sqrt{3}{)}^{n-3}\right(-1{)}^3$ + ...... + ${}^n{C}_n\left(\sqrt{3}{)}^0\right(-1{)}^n$

Clearly $x-I=2k$ for some k$\in I$ and also $0<y<1$

We can expand x as, x= $\left[x\right]+$ {x} where 0$\le${x}<1

So , $\left[x\right]+$$x$ $-y=2k$, since [x] is an integer, {x} - y must also be an integer as RHS is an integer .

So $x-y=0$ since $-1<x-y<1$

So. $\left[x\right]=2k$ for some $k\in I$