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KoMaL A Problems
KoMaL A Problems 2022/2023
A. 854
A. 854
Part of
KoMaL A Problems 2022/2023
Problems
(1)
Easy... just a long sum of tiny numbers
Source: KoMaL A. 854
6/13/2023
Prove that
∑
k
=
0
n
2
2
k
⋅
2
k
+
1
2
2
k
+
3
2
k
<
4
\sum_{k=0}^n\frac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}}<4
k
=
0
∑
n
2
2
k
+
3
2
k
2
2
k
⋅
2
k
+
1
<
4
holds for all positive integers
n
n
n
.Submitted by Béla Kovács, Szatmárnémeti
inequalities
algebra