[gon]

ตอนที่ 1.

1. $3^3 \times (-2)^4 = 432$

2. $a_n = \frac{1}{4}(\sqrt{(n+1)^2+n^2} - \sqrt{n^2+(n-1)^2}) $
$ \therefore s_{119} = \frac{1}{4}(\sqrt{120^2 + 119^2} - \sqrt{1^2+0^2}) = \frac{1}{4}(169-1) = 42$