# Thread: Sequence problem

1. ## Sequence problem

Determine the value for n for which

$(2^1)(2^2)(2^3)(2^4)...(2^n)=(2^210)$

I don't know what do here to be honest. Any help would be greatly appreciated, thanks!

EDIT: it's 2 to the power of n = 2 to the power of 210, if that's unclear.

2. Hello, Dovud!

Determine the value for $n$ for which:

. . $(2^1)(2^2)(2^3)(2^4)\:\cdots\:(2^n)\;=\;2^{210}$

We have: . $2^{(1+2+3+\cdots+n)} \;=\;2^{210}$

. . Hence: . $1 + 2 + 3 + \cdots + n \;=\;210$

The sum of the first $n$ natural numbers is given by: . $\frac{n(n+1)}{2}$

. . So we have: . $\frac{n(n+1)}{2} \:=\:210 \quad\Rightarrow\quad n^2 + n - 420 \:=\:0$

. . Factor: . $(n-20)(n+21) \:=\:0 \quad\Rightarrow\quad n \;=\;20,\;{\color{red}\rlap{/////}}-21$

Therefore: . ${\color{blue}n \;=\;20}$

3. Thank you!

4. EDIT: it's 2 to the power of n = 2 to the power of 210, if that's unclear.
To write more complicated exponents into LaTeX, insert them into braces :

2^{210} $\Rightarrow 2^{210}$
2^{3x + \sqrt{4}} $\Rightarrow 2^{3x + \sqrt{4}}$