# 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?

• This method is very efficient. Why would you want another ?
– user65203
Commented Jul 21, 2019 at 21:08

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 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 $$2^{10}=1024$$.

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 $$1024$$.

• You missed an ^ in your $2n-1$. It should be $2^n-1$ Commented Aug 15, 2019 at 17:54

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 $$1024$$.

• How did you get this $2^0+2^1+2^2+\dots+2^n=2^{n+1}-1$ ? Commented Jul 22, 2019 at 19:05
• @Kaushik In binary $\underbrace{11\dots1}_{n\text{ times}}+1=1{\underbrace{00\dots0}_{n\text{ times}}}$. Or use the standard formula $1+x+x^2+\dots+x^n=\frac{x^{n+1}-1}{x-1}$ (valid for $x\ne1$). Commented Jul 22, 2019 at 20:49
• But why is this true ?->[$2^0+2^1+2^2+\dots+2^n=2^{n+1}-1$] Commented Jul 22, 2019 at 21:22
• @Kaushik Doesn't the previous comment answer your question? Commented Jul 22, 2019 at 21:27
• @Kaushik The lhs is a number in binary representation Commented Jul 23, 2019 at 7:03

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

$$2^0 + 2^1 + 2^2 + 2^3 + ... + 2^n < 1025 = 2^{10}+ 2^0$$ $$2^1 + 2^2 + 2^3 + ... + 2^n < 2^{10}$$ $$\frac{1}{2^9} + \frac{1}{2^8} + ... + \frac{1}{2^{10-n}} < 1$$

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.

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, $$a_{1024}=\color{red}{a_{1025}}=\cdots =a_{2047}=\color{red}{1024}$$.