There are multiple answers to this question because a finite number of terms do not precisely determine the sequence. For example, one could fit a polynomial f(x) of degree 10 to this sequence so that f(i) was equal to the ith term.
However here is another approach
First lets let [x] be the greatest integer in x so that, for example
[1/2] = 0, [5/2] = 2, and so on. Note that if k is an even number [k/2] = k/2, and if k is an odd number [k/2] = (k-1)/2 , and  = 0.
I will assume the sequence starts with S0,
then the mth term in the sequence Sm is given by the formula
Sm = (2(m - 2[m/2]))∑[i≥0, i≤[m/2])2i
The way this formula works is that
(1) for even numbers m (including zero) the leading factor becomes 1, and for odd numbers it's 2.
(2) for even numbers the summation goes from 0 to m/2; for odd numbers it goes from 0 to (m-1)/2
So, for example, the 7th number in the sequence would be 2*(23+22+21+20) = 30;
and the 10th number in the sequence would be 1*(25+24+23+22+21+20) = 63
The sum of the powers of 2 from 0 to [m/2] can also be expressed as 2([m/2]+1)-1, so we could rewrite the formula as
Sm = (2(m - 2[m/2]))*(2([m/2]+1)-1)
So the 2010th term requested in the problem, since we defined the first term in the sequence to be S0, would be S2009
S2009 = 2*(21005-1) = 21006-2
I couldn't find any calculator that would compute this number without rounding. It's a number with about 303 digits.