2、Arrange seven distinct non-zero perfect squares in a row, such that the sum of any three consecutive numbers is greater than 100. The minimum value of the sum of these seven numbers is $\qquad$
2. 211

Analysis: Arrange the perfect squares $a_{1}, a_{2}, \ldots, a_{7}$ in a sequence. It is easy to see that: $a_{1}+a_{2}+a_{3}>100$, $a_{4}+a_{5}+a_{6}>100$. Therefore, the sum of these 7 numbers is at least 203. If the sum of 7 perfect squares is at least 203, they could be, in ascending order: $203=4+9+16+25+36+49+64$, $205=1+4+9+25+36+49+81$, $211=1+4+9+16+36+64+81$. We will now prove that 203 and 205 do not have valid constructions:
（1）When the sum of the 7 numbers is 203, because $a_{1}+a_{2}+a_{3}>100, a_{4}+a_{5}+a_{6}>100, a_{7} \geq 4$, it does not exist.
（2）When the sum of the 7 numbers is 205, if we select any 3 numbers such that their sum is greater than 100, then the sum of these three numbers is at least 107. Therefore, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}>107 \times 2+1=215$, which is a contradiction, so it does not exist.

Therefore, the minimum sum is 211, and a valid construction is: $1,36,64,9,81,16,4$.