A Pattern Might Emerge!

Algebra Level 4

$\large{1 + 2 \cdot (1 + 2 \cdot (1 + 2 \cdot ( \ldots \cdot (1 + 2) \ldots )))}$

There are 2016 pairs of parentheses in the expression above. What is the value of this expression?

2^{2016}-1

2^{2016}+1

2^{2017}-1

2^{2017}+1

2^{2018}-1

2^{2018}+1

2 solutions

Zk Lin
Feb 27, 2016

Denote $x_{1}=3$ , $x_{n+1}=2x_{n}+1$ for $n \geq 1$ . The problem is equivalent to finding $x_{2017}$ .

Observe that $x_{1}=3, x_{2}=7, x_{3}=15, x_{4}=31$ , this leads naturally to the conjecture that $x_{n}=2^{n+1}-1$ .

Substituting this into our recurrence equation $x_{n+1}=2x_{n}+1$ , we find that

L.H.S. $=2^{n+2}-1$

R.H.S. $=2(2^{n+1}-1)+1=2^{n+2}-1$

L.H.S. $=$ R.H.S., so the conjecture is true.

Therefore, $x_{2017}=\boxed{2^{2018}-1}$

Moderator note:

Great explanation. To keep to the presentation of the problem, it would be better to state that $x_{n+1} = 1 + 2 x_n$ instead, which follows immediately from seeing the substitution.

Ashish Menon
Apr 30, 2016

$\begin{aligned} 1 + 2 & = 3 & = 2^2 - 1\\ 1 + 2(1 + 2) & = 7 & = 2^3 - 1\\ 1 + 2 (1 + 2 (1 + 2)) & = 15 & = 2^4 - 1\\ 1 + 2 (1 + 2 (1 +2 (1 + 2))) & = 31 & = 2^5 - 1\\ 1 + 2 (1 + 2 (1 + 2 (1 +2 (1 +2)))) & = 63 & = 2^6 - 1 \end{aligned}$

So, we see that the expresson with $n$ parentheses = $2^{n + 2} - 1$

So, the expression with $2016$ parentheses $= 2^{2016 + 2} - 1\\ = \boxed{2^{2018} - 1}$

A Pattern Might Emerge!

2 solutions

Moderator note:

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