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2002 AIME Problems
7
Sum of First k Squares
Sum of First k Squares
Source:
December 28, 2006
Problem Statement
It is known that, for all positive integers
k
,
k,
k
,
1
2
+
2
2
+
3
2
+
⋯
+
k
2
=
k
(
k
+
1
)
(
2
k
+
1
)
6
.
1^{2}+2^{2}+3^{2}+\cdots+k^{2}=\frac{k(k+1)(2k+1)}{6}.
1
2
+
2
2
+
3
2
+
⋯
+
k
2
=
6
k
(
k
+
1
)
(
2
k
+
1
)
.
Find the smallest positive integer
k
k
k
such that
1
2
+
2
2
+
3
2
+
⋯
+
k
2
1^{2}+2^{2}+3^{2}+\cdots+k^{2}
1
2
+
2
2
+
3
2
+
⋯
+
k
2
is a multiple of
200.
200.
200.
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