Are all of them powers?

Algebra Level 2

$1, 1, 2, 4, 8, 16 , \ldots$

The above shows a sequence of integers with the first 2 terms of 1, and each subsequent terms is the sum of all the previous terms.

What is the $2017^\text{th}$ term of this sequence?

2^{2018}

2^{2017}

2^{2016}

2^{2015}

2 solutions

Sharky Kesa
Jan 21, 2017

Firstly, we will prove that the $n$ th term is twice the previous term for $n\geq 3$ . Let the $n$ th term of this sequence be $a_n$ .

$\begin{aligned} a_{n+1} &= \displaystyle \sum_{i=1}^{n} a_i\\ &= a_n + \displaystyle \sum_{i=1}^{n-1} a_i\\ &= a_n + a_n\\ &= 2a_n\\ \end{aligned}$

Also note that $a_3 = 2$ , so we must have $a_n = 2^{n-2}$ for $n \geq 2$ . Thus, $a_{2017} = 2^{2015}$ .

Toshit Jain
Mar 1, 2017

Looking at the sequence, we can conclude that 2^1=3rd term , 2^2=4th term . For any general 'n'th term :- 'n'th term = 2^(n-2). So, If n=2017 , the 2017th term will be 2^(2017-2) = 2^2015.

Are all of them powers?

2 solutions

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