Question

Given that k is a positive integer. Find, in terms of k, an expression for Sn, which is the sum of the integers from 2k to 4k inclusive​

Answer 1

Answer:

$S_n=3k\cdot (2k+1)=6k^2 +3k$

Step-by-step explanation:

k is a positive integer.

Consider an arithmetic sequence:

$a_1=2k\\ \\a_n=4k\\ \\d=1$

First, find n:

$a_n=a_1+(n-1)\cdot d\\ \\4k=2k+(n-1)\cdot 1\\ \\2k=n-1\\ \\n=2k+1$

Now, find the sum of these 2k+1 terms:

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\ \\S_n=\dfrac{2k+4k}{2}\cdot (2k+1)=\dfrac{6k}{2}\cdot (2k+1)=3k\cdot (2k+1)=6k^2 +3k$