Write the sum 1 + 4 + 7 + 10 + isdots + (3n-2) in sigma notation.

The terms are $1, 4, 7, 10, \dots$.
Calculate the difference between consecutive terms:
$4 - 1 = 3$
$7 - 4 = 3$
$10 - 7 = 3$
This is an arithmetic progression with a common difference $d = 3$.

The first term is $a_1 = 1$.
The formula for the $k$-th term of an A.P. is $a_k = a_1 + (k-1)d$.
Substitute $a_1 = 1$ and $d = 3$:

The sum starts with the first term where $k=1$, which gives $3(1)-2 = 1$.
The last term given is $(3n-2)$. Comparing this with the general term $3k-2$, we see that the upper limit for $k$ is $n$.