I need the steps used to figure out the equation.

## Determine the 2010th term for the following sequence: 1,2,3,6,7,14,15,30,31,62,63

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# 4 Answers

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.

Determine the 2010th term for the following sequence: 1,2,3,6,7,14,15,30,31,62,63

Let’s look at the odd terms, 1,3,7,…

k n t

1 0 1 = 1

3 1 3 = 2 + 1

5 2 7 = 4 + 2 + 1

7 3 15 = 8 + 4 + 2 + 1

…

k n t = 2^(n)+2^(n-1)+…+2+1

= 2^(n+1)-1, it’s a binary number and geometric series!

We want to know the 2010th number in the given sequence;

i.e., k = 2n+1 = 2010

2n = 2009

n = 2009/2 which isn’t evenly divisible,

so we want the even k term that’s 1 less than for k = 2011:

k = 2n+1 = 2011

2n = 2010

n = 1005

t = 2^(1005+1)-1

t-1 = 2^(1006)-2

So the 2010th number in the series is 2^(1006)-2.

Let’s look at the odd terms, 1,3,7,…

k n t

1 0 1 = 1

3 1 3 = 2 + 1

5 2 7 = 4 + 2 + 1

7 3 15 = 8 + 4 + 2 + 1

…

k n t = 2^(n)+2^(n-1)+…+2+1

= 2^(n+1)-1, it’s a binary number and geometric series!

We want to know the 2010th number in the given sequence;

i.e., k = 2n+1 = 2010

2n = 2009

n = 2009/2 which isn’t evenly divisible,

so we want the even k term that’s 1 less than for k = 2011:

k = 2n+1 = 2011

2n = 2010

n = 1005

t = 2^(1005+1)-1

t-1 = 2^(1006)-2

So the 2010th number in the series is 2^(1006)-2.

2^(1006) = 10^(1006 log(2)) ≈ 6.857655085992232 * 10^302

Here’s another construct where this sequence occurs:

2 | 1 1 1 1 1 1 ... 1

2 6 14 30 62 ... R-1

1 3 7 15 31 63 ... | R = p(2)

p(x) = x^n + x^(n-1) + x^(n-2) + … + x + 1

and p(2) is our binary number.

Here is a link to an article that points out a bug in Wolfram Alpha related to a comic on xkcd.

# Comments

Haha nice, I personally wouldn't trust any computer calculation out past ~20 digits, and there are very few applications that actually require that level of precision. I believe they calculate pi out to thousands of digits one at a time, therefore not
ever needing to store the entire number out to full precision.

This is a little hard to answer without knowing exactly what level of math this question is intended for. But, you can solve for the k

^{th}term via the recursion relation given byJ

_{k}= J_{k-1}+ 1 for odd kJ

_{k}= 2J_{k-1 }for even kwith J

_{0}= 0, and k>0using the above, we get

J

_{1}= 1J

_{2}= 2J

_{3}= 3J

_{4}= 6J

_{5}= 7J

_{6}= 14and so forth....

This recursive algorithm is trivial to implement numerically by writing a simple script in MATLAB, Python, or writing a simple C code. I get a value of 6.8577×10

^{302}(to 4 digits of precision).For a problem like this, there is no real trick to getting the equation other than staring at the numbers and trying to find out a pattern. Once you have the pattern, the trick is to figure out a way to express it mathematically, which again really just
sitting and thinking it through. The best way to get good at this type of problem is to just do as many as you can to get used to looking at sequences of numbers and finding patterns. This is a real case of practice makes perfect.

## 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.Comment