A problem for 6th graders

Algebra Level 2

$\large 4+2^2+2^3+2^4+\cdots+2^{20} = \ ?$

2^{18}

2^{21}

2^{19}

2^{22}

2^{20}

2 solutions

Thành Đạt Lê
Aug 17, 2017

This is far too easy for 6th graders. But, anyway,... $4 + 2^2 + 2^3 + 2^4 + ... + 2^{20}$

$= 2^2 + 2^2 + 2^3 + 2^4 + ... + 2^{20}$

$= 2^3 + 2^3 + 2^4 + ... + 2^{20}$

$= 2^4 + 2^4 + ... + 2^{20}$

$=...$

$= 2^{21}$ .

Mahdi Raza
Jun 29, 2020

\[\begin{align} E &= (1 + 2^0 + 2^1) + 2^2 + 2^3 + \ldots + 2^{20} \\ &= 1 + 2^0 + 2^1 + 2^2 + 2^3 + \ldots + 2^{20} \\ &= 1 + \dfrac{2^0(2^{21} - 1)}{2-1} \\ &= 1 + 2^{21} - 1 \\ &= \boxed{2^{21}}

\end{align}\]

A problem for 6th graders

2 solutions

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