Given the sequence #2, 6, 12, 20,...# what is the formula for the #n#th term and for the sum to #n# terms ?

Given the sequence #2, 6, 12, 20#, write it out in a line with gaps between each term:

#color(blue)(2)color(white)(00000)color(green)(6)color(white)(00000)12color(white)(00000)20#

Add a line of term below listing the differences between each pair of terms:

#color(blue)(2)color(white)(00000)color(green)(6)color(white)(00000)12color(white)(00000)20#
#color(white)(000)color(blue)(4)color(white)(00000)color(green)(6)color(white)(000000)8#

Add another line of term below listing the differences between each pair of terms:

#color(blue)(2)color(white)(00000)color(green)(6)color(white)(00000)12color(white)(00000)20#
#color(white)(000)color(blue)(4)color(white)(00000)color(green)(6)color(white)(000000)8#
#color(white)(000000)color(blue)(2)color(white)(000000)color(green)(2)#

Notice that both terms of the last line are the same.

That implies that our given sequence can be matched using a quadratic formula, which we can construct using the first term of each of the lines as a coefficient:

#a_n = color(blue)(2)/(0!) + color(blue)(4)/(1!)(n-1) + color(blue)(2)/(2!)(n-1)(n-2)#

#color(white)(a_n) = 2+4n-4+n^2-3n+2#

#color(white)(a_n) = n^2+n#

If you know the formulas for #sum_(n=1)^N n^2# and #sum_(n=1)^N n# then you can just add them to provide the formula for #sum_(n=1)^N a_n#.

I would rather construct it directly:

Add another line to the top of our sequences, consisting of the sums of the terms on the original top line:

#color(blue)(0)color(white)(00000)color(green)(2)color(white)(00000)8color(white)(00000)20color(white)(00000)40#
#color(white)(000)color(blue)(2)color(white)(00000)color(green)(6)color(white)(00000)12color(white)(00000)20#
#color(white)(000000)color(blue)(4)color(white)(00000)color(green)(6)color(white)(000000)8#
#color(white)(000000000)color(blue)(2)color(white)(000000)color(green)(2)#

We want to start with #s_1 = 2#. So ignore the first terms of each sequence and use the second terms to give us our formula for the sum to #n# terms:

#s_n = color(green)(2)/(0!)+color(green)(6)/(1!)(n-1)+color(green)(6)/(2!)(n-1)(n-2) + color(green)(2)/(3!)(n-1)(n-2)(n-3)#

#color(white)(s_n) = 2+6n-6+3n^2-9n+6+1/3n^3-2n^2+11/3n-2#

#color(white)(s_n) = 1/3(n^3+3n^2+2n)#

#color(white)(s_n) = 1/3n(n+1)(n+2)#