2. Twenty numbers are arranged in a sequence. It is known that each number in the sequence, except the first one, is twice as large as the number immediately preceding it. How many times smaller is the sum of the first ten numbers in the sequence compared to the sum of the last ten numbers in the sequence?
2. Twenty numbers are arranged in a sequence. It is known that each number in the sequence, except the first, is twice as large as the number immediately preceding it. How many times smaller is the sum of the first ten numbers in this sequence compared to the sum of the last ten numbers in this sequence?

## Solution.

The numbers in the sequence are:

$x, 2 x, 2^{2} x, 2^{3} x, \ldots, 2^{18} x, 2^{19} x$.

1 POINT

The sum of the first ten terms of the sequence:

$x+2 x+2^{2} x+2^{3} x+2^{4} x+2^{5} x+2^{6} x+2^{7} x+2^{8} x+2^{9} x=x\left(1+2+2^{2}+\ldots+2^{9}\right)$

1 POINT

The sum of the last ten terms of the sequence:

$2^{10} x+2^{11} x+2^{12} x+2^{13} x+2^{14} x+2^{15} x+2^{16} x+2^{17} x+2^{18} x+2^{19} x$

$=2^{10} \cdot x\left(1+2+2^{2}+\ldots+2^{9}\right)$

2 POINTS

$k \cdot x\left(1+2+2^{2}+\ldots+2^{9}\right)=2^{10} \cdot x\left(1+2+2^{2}+\ldots+2^{9}\right)$

$k=2^{10}$

$k=1024$

2 POINTS

The sum of the first ten numbers in this sequence is 1024 times smaller than the sum of the last ten numbers in this sequence. TOTAL 6 POINTS

Note 1: If the student does not calculate that $2^{10}=1024$, they should receive (at most) 5 POINTS.

Note 2: If the student assumes $s x=1$, they should lose 1 POINT (thus, a student who solves the problem this way to completion should receive 5 POINTS).

Note 3: If the student calculates all powers of 2 up to $2^{19}$ and then adds and divides them without factoring out the common factor and makes a mistake in the calculation or does not complete the problem to the end, they can receive a maximum of 3 POINTS (whether they receive 1, 2, or 3 POINTS depends on how much calculation they have done and how many mistakes they have made).