Answer
$a_n=n(n+1)$
Work Step by Step
We have to write an expression for the general term of sequence:
$2,6,12,20,30, . . .$
Let's find the pattern for the general term:
$a_2-a_1=6-2=4=2+2\cdot1,$ $a_2=a1+2 +2\cdot1$
$a_3-a_2=12-6=6=2+2\cdot2,$ $a_3=a_2+2+2\cdot2$
$a_4-a_3=20-12=8=2+2\cdot3,$ $a_4=a_3+2+2\cdot3$
$a_5-a_4=30-20=10=2+2\cdot4,$ $a_5=a_4+2+2\cdot4$
$a_1=2$
$a_2=a_1+2+2\cdot1$
$a_3=a_1+2+2\cdot1+2+2\cdot2=a_1+2\cdot2+2(1+2)$
$a_4=a_1+2+2\cdot1+2+2\cdot2+2+2\cdot3$
$a_4=a_1+2\cdot3+2(1+2+3)$
$a_5=a_1+2+2\cdot1+2+2\cdot2+2+2\cdot3+2+2\cdot4$
$a_5=a_1+2\cdot4+2(1+2+3+4)$
$a_n=a_1+2(n-1)+2(1+2+3+ . . . +(n-1))$
The value in brackets is an arithmetic progression.
$a_n=a_1+2(n-1)+2\dfrac{1+(n-1)}{2}(n-1)$
$$a_n=2+2(n-1)+n(n-1)=n^2+n=n(n+1)$$
Another solution:
We can write:
$a_1=2=1\cdot 2=1\cdot (1+1)$
$a_2=6=2\cdot 3=2\cdot (2+1)$
$a_3=12=3\cdot 4=3\cdot (3+1)$
$a_4=20=4\cdot 5=4\cdot (4+1)$
So the general term is:
$a_n=n(n+1)$
