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Day 1, Problem 2
IMC 2010 · Problem 2
Day 1
17th IMC · Blagoevgrad, Bulgaria
Statement
Compute the sum of the series
∑
k
=
0
∞
1
(
4
k
+
1
)
(
4
k
+
2
)
(
4
k
+
3
)
(
4
k
+
4
)
=
1
1
⋅
2
⋅
3
⋅
4
+
1
5
⋅
6
⋅
7
⋅
8
+
…
.
\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \dots.
k
=
0
∑
∞
(
4
k
+
1
)
(
4
k
+
2
)
(
4
k
+
3
)
(
4
k
+
4
)
1
=
1
⋅
2
⋅
3
⋅
4
1
+
5
⋅
6
⋅
7
⋅
8
1
+
…
.
Official solution
Reveal
Hidden so you can work on the problem first.
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