Question

find the \(\large 1031^{th}\) term of the sequence.

Answer 1

\(\large \color{black}{\begin{align}

1,\ 2,\ 2,\ 4,\ 4,\ 4,\ 4,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\cdots \ \infty +\hspace{.33em}\\~\\


\end{align}}\)

Answer 2

You have already shown:
1st term=1
3rd term=2
7th term=4
15th term=8
...
can you find the
1023rd term?
and hence the 1031st term?

Answer 3

no i didnt understand

Answer 4

1024

Answer 5

or 512

Answer 6

In case you haven't found it yet:

1st term=\(2^1-1\) st term=1 =\(2^0\)
3rd term=\(2^2-1\) rd term=2=\(2^1\)
7th term=\(2^3-1\) th term=4=\(2^2\)
15th term=\(2^4-1\) th term=8=\(2^3\)
...
can you find the
1023rd term? =\(2^{10}-1\) rd term = \(2^9\) = 512
and hence the 1031st term?

Answer 7

is it 1024

Answer 8

Yes, the next one is 1024! In fact, the 1024th term starts at 1024, etc.