The first term of a number sequence is 1. Each subsequent term is 4 times the square root of the sum of the preceding terms. What is the sum of the first 1971 terms of the sequence?
We denote the $n$-th term of the sequence by $a_{n}$, so $a_{1}=1$ and for $n=2,3,4, \ldots$

$$
a_{n}=4+4 \sqrt{a_{1}+a_{2}+\ldots+a_{n-1}}
$$

Accordingly, $a_{2}=4+4 \sqrt{1}=8, a_{3}=4+4 \sqrt{1+8}=16, a_{4}=4+4 \sqrt{1+8+16}=24, a_{5}=32, a_{6}=40$. During the calculations, the results of the consecutive square roots were $1,3,5,7,9$, which are the odd numbers one less than the index for $n=2,3,4,5,6$. We will prove that our observation holds for every index.

Suppose $k$ is an index for which the sum of the first $k$ terms of the sequence, $s_{k}$, is equal to the square of the $k$-th odd number, that is,

$$
s_{k}=a_{1}+a_{2}+\ldots+a_{k}=(2 k-1)^{2}
$$

Then, according to (1),

$$
\begin{aligned}
a_{k+1}=4+4(2 k-1) & =8 k, \quad \text { and thus } \\
s_{k+1}=s_{k}+a_{k+1}=(2 k-1)^{2}+8 k & =4 k^{2}+4 k+1=(2 k+1)^{2}
\end{aligned}
$$

so the stated property also holds for the next $(k+1)$ index. This is what we wanted to prove.

According to this, the sum of the first 1971 terms of the sequence is the square of the 1971st odd number:

$$
s_{1971}=(2 \cdot 1971-1)^{2}=3941^{2}=15531481
$$

Péter Erdős (Budapest, I. István Gymn., II. o. t.)