The nth term of a series is denoted by $(-1)^n 3^{n-1}+2n$. What is the sum of the first six terms of this series?

I plugged in n=1 through 6 and got the sum. The alternating power of 3 is a bit tricky to compute, but doable. What did you all get?

For the term (-1)^n 3^{n-1}, when n is odd, it's negative; when even, positive. So the first six terms: n=1: -1 +2=1; n=2: 3+4=7; n=3: -9+6=-3; n=4: 27+8=35; n=5: -81+10=-71; n=6: 243+12=255. Sum = 1+7-3+35-71+255 = 224? Wait, let me recalc: 1+7=8, 8-3=5, 5+35=40, 40-71=-31, -31+255=224. That's option D. But I thought the answer was 194? Hmm, maybe I made a mistake. Let me check n=2: (-1)^2=1, 3^{1}=3, 2n=4, so 3+4=7. n=4: (-1)^4=1, 3^3=27, 2n=8, 27+8=35. n=6: (-1)^6=1, 3^5=243, 2n=12, 243+12=255. Sum: 1+7=8, 8-3=5, 5+35=40, 40-71=-31, -31+255=224. Yes, 224. So answer is D. But earlier someone said 194? That's for a different sum maybe.

This one is a bit calculation heavy. I like to separate the sum into the alternating geometric part and the arithmetic part. Sum of (-1)^n 3^{n-1} for n=1..6 is -1+3-9+27-81+243 = 182? Let's see: -1+3=2, 2-9=-7, -7+27=20, 20-81=-61, -61+243=182. And sum of 2n for n=1..6 is 2*(1+2+3+4+5+6)=2*21=42. Total 182+42=224. So D. But wait, 182 is an answer choice, so maybe some forget to add the 2n part?