Fun with 2018 #1

Algebra Level 3

$\large \begin{aligned} & 1 + 4 + 12 + 32 + \ldots + 2018 \cdot 2^{2017} \\ & = 1 \cdot 2^0 + 2 \cdot 2^1 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 2018 \cdot 2^{2017} \\ & = \ ? \end{aligned}$

-1 + 2018 \cdot 2^{2018}

-1 + 2017 \cdot 2^{2018}

1 + 2018 \cdot 2^{2018}

1 + 2017 \cdot 2^{2018}

1 solution

Yash Ghaghada
Jan 31, 2018

Fun with 2018 #1

1 solution

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