# Problem 3. (3 points)

The sequence $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=2$ and $a_{n}=a_{1}+a_{2}+\ldots+a_{n-1}+n$ for $n \geqslant 2$. Find the explicit formula for this sequence.
Answer: $a_{n}=5 \cdot 2^{n-2}-1$ for $a>1, a_{1}=2$

Solution:

By subtracting the formula for $a_{n}$ from the formula for $a_{n+1}$, we get $a_{n+1}-a_{n}=a_{n}+1$, that is, $a_{n+1}=2 a_{n}+1$ for $n>1$.

Now let's prove the formula $a_{n}=5 \cdot 2^{n-2}-1$ by induction. The base case is $n=2$, and indeed, $a_{2}=a_{1}+2=4=5 \cdot 2^{0}-1$.

Next, we will use the obtained recurrence formula and verify that if $a_{n}=$ $5 \cdot 2^{n-2}-1$, then $a_{n+1}=2\left(5 \cdot 2^{n-2}-1\right)+1=5 \cdot 2^{n-1}-1$.