Question

sum_{n=1}^{12}3n-2

Answer 1

sum_{n=1}^{12}3n-2

Answer 2

\[\sum_{n=1}^{12}3n-2\]

Answer 3

This is an arithmetic progression.

Answer 4

oops

Answer 5

\[1+4+7+10+...+34\]

Answer 6

what is the formula to figure this out

Answer 7

The of the first n terms of an arithmetic progression is\[S _{n}=\frac{ n }{ 2 }\left[ 2a +(n -1)d \right]\]where a is the first term and d is the common difference.

Answer 8

OR you can also use \[S _{n}=\frac{ n }{ 2 }(a +t _{n})\]where tn is the last term.

Answer 9

Since @satellite73 already gave it away, you know that the first term is 1 and the last term is 34. You also know that the common difference is 3 because each term of the progression increases by 3.

Answer 10

Now you just need to use one of the formulas above. The second would be more efficient in this case. But either is fine.

Answer 11

can you give me an example i'm still confused. i learned this about two years ago and i was taught through another formula that i dont remember

Answer 12

Sure. If you're given\[\sum_{n =1}^{11}(5n +3)\]for example. Then 5(1) + 3= 8 is the first term and the last term would be 5(11) + 3 = 58. Thus\[S _{11}=\frac{ 11 }{ 2 }(8+58)=363\]∴ the sum is 363. Do you understand?

Answer 13

Sorry for the delay. I was helping someone else, plus my server is sometimes slow, plus I take my time typing.

Answer 14

its okay thank you