Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 1"
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==Solution== | ==Solution== | ||
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| + | Note that for every positive integer <math>n > 2</math>, by the given recurrence relation, we have | ||
| + | \begin{align*} | ||
| + | a_n &= a_{n-1} + a_{n-2} + \cdots + a_1 \\ | ||
| + | &= (a_{n-2} + \cdots + a_1) + a_{n-2} + \cdots + a_1 \\ | ||
| + | &= 2(a_{n-2} + \cdots + a_1) \\ | ||
| + | &= 2a_{n-1}. | ||
| + | \end{align*} | ||
| + | Therefore, <math>a_{25} = 2a_{24} = 4a_{23} = 8a_{22} = 16a_{21} = 32a_{20} = \boxed{800}</math>. | ||
| + | |||
| + | ~Sedro | ||
Latest revision as of 01:54, 11 September 2025
Problem
An infinite sequence of real numbers 
satisfies ![\[a_n = a_1+a_2+\dots+a_{n-1}\]](http://latex.artofproblemsolving.com/7/d/f/7df1014ff9d256094d26e0549129f25d3d0734c1.png)
for all positive integers 
. Given that 
find 
.
Solution
Note that for every positive integer 
, by the given recurrence relation, we have
\begin{align*}
a_n &= a_{n-1} + a_{n-2} + \cdots + a_1 \\
&= (a_{n-2} + \cdots + a_1) + a_{n-2} + \cdots + a_1 \\
&= 2(a_{n-2} + \cdots + a_1) \\
&= 2a_{n-1}.
\end{align*}
Therefore, 
.
~Sedro
