The partial sum $1+5+9+\cdots+141$ equals

Step 1 ：This is an arithmetic series where the common difference is 4. The formula for the sum of an arithmetic series is given by: \(S_n = \frac{n}{2} (a_1 + a_n)\) where \(S_n\) is the sum of the first n terms, \(a_1\) is the first term, \(a_n\) is the nth term, and n is the number of terms.

Step 2 ：First, we need to find the number of terms in the series. We know that the first term \(a_1\) is 1 and the last term \(a_n\) is 141. We can use the formula for the nth term of an arithmetic series: \(a_n = a_1 + (n - 1) * d\) where d is the common difference. In this case, d is 4. We can solve this equation for n to find the number of terms in the series.

Step 3 ：Then, we can substitute the values of \(a_1\), \(a_n\), and n into the formula for the sum of an arithmetic series to find the sum.

Step 4 ：Given values: \(a_1 = 1\), \(a_n = 141\), \(d = 4\), \(n = 36\)

Step 5 ：Substituting these values into the formula, we get \(S_n = 2556\)

Step 6 ：Final Answer: The partial sum \(1+5+9+\cdots+141\) equals \(\boxed{2556}\)