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**TOUGH QUESTION**
For any n >= 1, the sum Sn of the first n terms in a certain arithmetic sequence
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Oh, I'm getting \(6n-4\)
Haha, gotta love that constant term...
\(S_n=n/2(2a+(n-1)d)=n^2(d/2)+n(a-d/2)=3n^2-n\) so comparing the co-efficients,we get d=3*2=6 a=2 so \(T_n=a+(n-1)d=2+6n-6=6n-4\)
yup,its 6n-4
u got that @artofspeed ?
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Here's how I did it: \[ \sum_{i=1}^{n}ai+b=3n^2-n\implies\\ \sum_{i=1}^{n} ai+\sum_{i=1}^{n}b=\frac{an(n+1)}2+bn=3n^2-n\implies\\ an^2+an+2bn=6n^2-2n\implies a=6, b=-4 \]Therefore: \[ A_n=an+b=6n-4 \]
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