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All-Russian Olympiad
1962 All Russian Mathematical Olympiad
015
ASU 015 All Russian MO d 1962 8.3 a_{100}=3a_{99}-2a_{98}
ASU 015 All Russian MO d 1962 8.3 a_{100}=3a_{99}-2a_{98}
Source:
June 17, 2019
algebra
Problem Statement
Given positive numbers
a
1
,
a
2
,
.
.
.
,
a
99
,
a
100
a_1,a_2,...,a_{99},a_{100}
a
1
,
a
2
,
...
,
a
99
,
a
100
. It is known, that
a
1
>
a
0
,
a
2
=
3
a
1
−
2
a
0
,
a
3
=
3
a
2
−
2
a
1
,
.
.
.
,
a
100
=
3
a
99
−
2
a
98
a_1>a_0, a_2=3a_1-2a_0, a_3=3a_2-2a_1, ..., a_{100}=3a_{99}-2a_{98}
a
1
>
a
0
,
a
2
=
3
a
1
−
2
a
0
,
a
3
=
3
a
2
−
2
a
1
,
...
,
a
100
=
3
a
99
−
2
a
98
Prove that
a
100
>
2
99
.
a_{100}>2^{99}.
a
100
>
2
99
.
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