Express the series 12 + 22 + 32 + s + 92 using sigma notation.

The terms in the series are squares of consecutive integers: $1^2, 2^2, 3^2, \dots, 9^2$.
The general term can be represented as $k^2$, where $k$ is an integer.

The series starts with $1^2$, so the lower limit of the index $k$ is 1.
The series ends with $9^2$, so the upper limit of the index $k$ is 9.
Thus, the sigma notation is $\sum_{k=1}^{9} k^2$.

Let's examine the options:

• (a) $\sum_{k=0}^{9} k^2$: This sum is $0^2 + 1^2 + 2^2 + \dots + 9^2$. Since $0^2 = 0$, this is equivalent to $1^2 + 2^2 + \dots + 9^2$.
• (b) $\sum_{k=1}^{9} (k+1)^2$: This sum is $2^2 + 3^2 + \dots + 10^2$.
• (c) $\sum_{k=1}^{9} k+2$: This sum is $(1+2) + (2+2) + \dots + (9+2) = 3+4+\dots+11$.
• (d) $\sum_{k=1}^{9} (k+2)^2$: This sum is $3^2 + 4^2 + \dots + 11^2$.

Option (a) correctly represents the given series because adding $0^2$ does not change the sum.

The final answer is $\boxed{\text{a}}$.