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2018 CHMMC (Fall)
3
2018 CHMMC Tiebreaker 3 - sum ( 1/(n^2 + 3n) - 1/(n^2 + 3n + 2) )
2018 CHMMC Tiebreaker 3 - sum ( 1/(n^2 + 3n) - 1/(n^2 + 3n + 2) )
Source:
March 2, 2024
algebra
Sum
CHMMC
Problem Statement
Compute
∑
n
=
1
∞
(
1
n
2
+
3
n
−
1
n
2
+
3
n
+
2
)
\sum^{\infty}_{n=1} \left( \frac{1}{n^2 + 3n} - \frac{1}{n^2 + 3n + 2}\right)
n
=
1
∑
∞
(
n
2
+
3
n
1
−
n
2
+
3
n
+
2
1
)
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