Answer:
[tex]S_n=3k\cdot (2k+1)=6k^2 +3k[/tex]
Step-by-step explanation:
k is a positive integer.
Consider an arithmetic sequence:
[tex]a_1=2k\\ \\a_n=4k\\ \\d=1[/tex]
First, find n:
[tex]a_n=a_1+(n-1)\cdot d\\ \\4k=2k+(n-1)\cdot 1\\ \\2k=n-1\\ \\n=2k+1[/tex]
Now, find the sum of these 2k+1 terms:
[tex]S_n=\dfrac{a_1+a_n}{2}\cdot n\\ \\S_n=\dfrac{2k+4k}{2}\cdot (2k+1)=\dfrac{6k}{2}\cdot (2k+1)=3k\cdot (2k+1)=6k^2 +3k[/tex]
