1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $

Solution 1:

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.

$$1025_{10}=‭10000000001_b\to‭10000000000_b=1024_{10}.$$‬

Solution 2:

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 $2^{10}=1024$.

Solution 3:

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 $1024$.

Solution 4:

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 $1024$.

Solution 5:

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 $4^{k} = 4^{5} = 1024.$