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 1995 terms of the sequence?

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Translation:
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 1995 terms of the sequence?
Let $a_{n}$ denote the $n$-th term of the sequence, and $s_{n}$ the sum of the first $n$ terms. By calculating the first few terms: $a_{1}=1, a_{2}=8, a_{3}=16$, and $s_{1}=1, s_{2}=9, s_{3}=25$. We will show that in general $s_{n}=(2 n-1)^{2}$. As we have seen, this is true for $n=1,2,3$.

Suppose that for some positive integer $k$ our statement is true, i.e., $s_{k}=(2 k-1)^{2}$. Then, based on the definition of the sequence, $a_{k+1}=4 \sqrt{s_{k}}+4=8 k$, and $s_{k+1}=s_{k}+a_{k+1}=(2 k-1)^{2}+8 k=$ $(2 k+1)^{2}=(2(k+1)-1)^{2}$, so the statement is also true for $n=k+1$.

In the case of $n=1995$, the sum of the first 1995 terms is $(2 \cdot 1995-1)^{2}=15912121$.