1. The problem is to find the sum of the series: 1 + 3 + 5 + 7 + ... + 69. 2. This is an arithmetic series where the first term $a_1 = 1$, the common difference $d = 2$, and the last term $a_n = 69$. 3. To find the number of terms $n$, use the formula for the $n$th term of an arithmetic sequence: $$a_n = a_1 + (n-1)d$$ Substitute the known values: $$69 = 1 + (n-1) \times 2$$ 4. Simplify and solve for $n$: $$69 - 1 = 2(n-1)$$ $$68 = 2(n-1)$$ $$n-1 = \frac{68}{2} = 34$$ $$n = 35$$ 5. The sum $S_n$ of the first $n$ terms of an arithmetic series is given by: $$S_n = \frac{n}{2} (a_1 + a_n)$$ 6. Substitute $n=35$, $a_1=1$, and $a_n=69$: $$S_{35} = \frac{35}{2} (1 + 69) = \frac{35}{2} \times 70$$ 7. Calculate the sum: $$S_{35} = 35 \times 35 = 1225$$ Therefore, the sum of the series 1 + 3 + 5 + ... + 69 is **1225**.