2000 AMC 10 Problems/Problem 12

Problem

We have a recursion:

$A_n=A_{n-1}+4(n-1)$ .

I.E. we add increasing multiples of $4$ each time we go up a figure.

So, to go from Figure 0 to 100, we add

$4 \cdot 1+4 \cdot 2+...+4 \cdot 99=4 \cdot 4950=19800$ .

$19801$

B.

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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