1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it  We can observe the following -

$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$ 
$5 - 4$
. . .
$8 - 8$ 
$9 - 8$
. . .

We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ . 
Is there any other method to solve this question?
 A: The number $n$ first appears in the sequence at position $n$ until position $2n-1$. So, the number $1024$ appears in the sequence at position $1024$ until $2047$. Therefore the number will be $1024$.
A: In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,\cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,\cdots$$
So for any index, clear all bits but the most significant.
$$1025_{10}=‭10000000001_b\to‭10000000000_b=1024_{10}.$$‬
A: A higher brow way of writing the same thing is to say the $n^{th}$ term is 
$2^{\lfloor \log_2 n\rfloor}$, then to evaluate that at $n=1025$.  The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^{10}=1024$.
A: The first term is $2^0=1$; the next $2^1=2$ terms are equal to $2^1=2$; the next $2^2=4$ terms are equal to $4$ and so on. Since
$$
2^0+2^1+2^2+\dots+2^n=2^{n+1}-1
$$
the term at place $1023$ is $512$. The next $2^{10}$ terms are equal to $1024$.
A: The given sequence is equivalent to
$$ 1 + 2(2) + 4(2^{2}) + 8(2^{3}) + 16(2^{4}) + \ldots $$
$$ = 1 + (2^{2})^{1} + (2^{2})^{2} + (2^{3})^{3} + \ldots + (2^{2})^{k-1} + \ldots $$
Now, since this is a geometric series, we may solve  
$$ s_{k} = \frac{a(1 - r^{k})}{1-r} = \frac{1(1-4^{k})}{1-4} = 1025 $$
which easily can be shown to give $ 5 < k < 6$.
Hence, the integer $k$ we seek is $5$; and so, the $1025th$ term of the sequence is $4^{k} = 4^{5} = 1024.$
A: $$ 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^n < 1025 = 2^{10}+ 2^0 $$
$$ 2^1 + 2^2 + 2^3 + ... + 2^n < 2^{10} $$
$$ \frac{1}{2^9} + \frac{1}{2^8} + ... + \frac{1}{2^{10-n}} < 1 $$
In the left-hand side of the above inequality, we've acquired a geometrical sequence. The summation formula for geometrical series is;
$$ S = \frac{a(1-r^n)}{1-r} $$
So this summation must be smaller than 1, with this condition stated we can set n = 9. 
$$ 2^0 + 2^1 + 2^2 + ... + 2^9 = 1023 $$
We are now able to conclude that the term at place 1024 and 1025 (at the same time) is 1024.
A: Make up the frequency and cumulative frequency table:
$$\begin{array}{c|c|c}
x&f&F\\
\hline
1&1&1=2^1-1\\
2&2&3=2^2-1\\
4&4&7=2^3-1\\
8&8&15=2^4-1\\
\vdots&\vdots&\vdots\\
256&256&511=2^{9}-1\\
512&512&1023=2^{10}-1\\
1024&1024&2047=2^{11}-1\\
\vdots&\vdots&\vdots\\
2^n&2^n&2^{n+1}-1
\end{array}$$
So, your approach was efficient to notice that $a_{2^n}=2^n$. Hence, $a_{1024}=\color{red}{a_{1025}}=\cdots =a_{2047}=\color{red}{1024}$.
