Question 934667
<pre>
I think the "2n terms" should have been "3n terms", because with 
2n terms, there will have to be three formulas, depending on whether 
2n is a multiple of 3, 1 less than a multiple of 3, or 1 more than 
a multiple of 3.

So I'm going to do the problem assuming that the 2n should have been 3n,

OK?  Let me know.

It's the sum of two arithmetic sequences:

S1 = 2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+...+3n+1 (to 3n terms) 
S2 =    -4    -7    -10      -13      -16+...      (to n terms   

There are 3n terms of the first sequence and n terms of the second sequence.

We use the formula:

{{{S[n]=expr(n/2)(a[1]+a[n])}}}

We substitute {{{3n}}} for {{{n}}}, {{{a[1]=2}}}, {{{a[3n]=3n+1}}}

{{{S1[3n]=expr(3n/2)(2+3n+1)}}}
{{{S1[3n]=expr(3n/2)(3n+3)}}}
{{{S1[3n]=expr(3n/2)(3(n+1))}}}
{{{S1[3n]=expr(9n/2)(n+1)}}} 
{{{S1[3n]=9n(n+1)/2}}} = the sum of the 1st sequence.

We use the formula 

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

to find the sum of the 2nd sequence

{{{a[1]=-4}}}, {{{d=-3}}} 

{{{S2[n]=expr(n/2)(2(-4)+(n-1)(-3))}}}
{{{S2[n]=expr(n/2)(-8-3n+3))}}}
{{{S2[n]=expr(n/2)(-5-3n))}}}
{{{S2[n]=n(-5-3n))/2}}}
{{{S2[n]=-n(5+3n))/2}}}

So the required sum is

{{{S1[3n]-S1[n]}}}{{{""=""}}}{{{9n(n+1)/2-(n(5+3n))/2}}}{{{""=""}}}

{{{(9n(n+1)-n(5+3n))/2}}}{{{""=""}}}{{{(9n^2+9n-5n-3n^2)/2}}}{{{""=""}}}

{{{6n^2+4n)/2}}}{{{""=""}}}{{{3n^2+2n}}}{{{""=""}}}{{{n(3n+2)}}}

Sum of the first 3n terms = {{{n(3n+2)}}}

Edwin</pre>