Geometric series, eh?

The sum to infinity of a geometric series exists only if

$|r| < 1$ , where $r$ is the common ratio of the geometric series.

Let us assume $u_1 = a , a \in \mathbb{R}$ and the common ratio to be $r$ .

$\displaystyle \sum_{i=1}^{\infty} u_i = 3 = \dfrac{a}{1-r}$

$\displaystyle \sum_{i=1}^{\infty} u_{i}^{2} = 3 =\dfrac{a^2}{1-r^2}$

$\Rightarrow 9 = \dfrac{a^2}{(1-r)^2}$

$\Rightarrow 9 = \dfrac{3(1-r^2)}{(1-r)^2}$

Rearranging the terms, we get

$4r^2 - 6r + 2 = 0 \Rightarrow r = \dfrac{1}{2} , 1$

But $r \neq 1$ because series will not converge.

$\Rightarrow r = \dfrac{1}{2}$

Substituting this in the first equation, we get $a = \dfrac{3}{2}$

Hence,

$u_n = ar^{n-1} = \dfrac{3}{2}\left(\dfrac{1}{2}\right)^{n-1} = \dfrac{3}{2^n}$

$u_{10} = \dfrac{3}{1024} \Rightarrow \lfloor 1000u_{10} \rfloor = \boxed{2}$

Let the geometric series be $u_i=ar^{i-1}$ . Since the series is convergent and not zero, it is safe to assume that $a\ne 0, r \ne \pm 1$ .

Now, we have

$\sum_{i=1}^{\infty} u_i = \frac{a}{(1-r)}=3 \ ....(A)$

which implies $a=3(1-r)$ ....(B)

and

$\sum_{i=1}^{\infty} u_i^2 = \frac{a^2}{(1-r^2)}=3\ ....(C)$

Dividing (C) by (A),

$\frac{a^2}{(1-r^2)}\frac{(1-r)}{a}=\frac{a}{(1+r)}=1$

This gives $a=(1+r)$ .....(D)

Using, (B) and (D), we get $a=1.5,r=0.5$ .

The required result is

$\lfloor 1000ar^9\rfloor = \lfloor 1000 \frac{3}{2}\frac{1}{512} \rfloor = \lfloor \frac{3000}{1024}\rfloor = \boxed{2}$

It should be specified about the greatest integer function !!