I need the steps used to figure out the equation.

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] = 0.

I will assume the sequence starts with S

_{0},then the m

^{th}term in the sequence S_{m}is given by the formulaS

_{m}= (2^{(m - 2[m/2])})∑_{[i≥0, i≤[m/2])}2^{i}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*(2

^{3}+2^{2}+2^{1}+2^{0}) = 30;and the 10th number in the sequence would be 1*(2

^{5}+2^{4}+2^{3}+2^{2}+2^{1}+2^{0)}= 63The 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 asS

_{m}= (2^{(m - 2[m/2])})*(2^{([m/2]+1)}-1)===========================

So the 2010

^{th}term requested in the problem, since we defined the first term in the sequence to be S_{0}, would be S_{2009}S

_{2009}= 2*(2^{1005}-1) = 2^{1006}-2I couldn't find any calculator that would compute this number without rounding. It's a number with about 303 digits.

## Comments

^{1006}= 2^{1004}*2^{4}, and 1004/4 = 251. This means that 2^{1006}ends in the digit 4, which means that the Wolfram answer ends with the correct digit.