Verified answer

[tex]\displaystyle\bf\\a_{1} =1\\\\d=5\\\\\\S_{n} =\frac{2a_{1} +d\cdot(n-1)}{2} \cdot n=\frac{2\cdot 1+5\cdot (n-1)}{2} \cdot n=\\\\\\\frac{2+5n-5}{2} \cdot n=\frac{5n-3}{2}\cdot n\\\\\\Otvet: \ S_{n} =\frac{5n-3}{2} \cdot n[/tex]