To find the sum of coefficients of odd powers of x in the expansion of (1 + x)^60, we can use the Binomial Theorem.

The Binomial Theorem states that for any positive integer n, the expansion of (a + b)^n can be written as:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

Where C(n, r) represents the binomial coefficient, which is the number of ways to choose r items from a set of n items.

In this case, we have (1 + x)^60, so a = 1 and b = x. Plugging these values into the Binomial Theorem, we get:

(1 + x)^60 = C(60, 0) * 1^60 * x^0 + C(60, 1) * 1^59 * x^1 + C(60, 2) * 1^58 * x^2 + ... + C(60, 59) * 1^1 * x^59 + C(60, 60) * 1^0 * x^60

Now, let's focus on the terms with odd powers of x. These are the terms where the exponent of x is odd, meaning the power of x is either 1, 3, 5, ..., or 59.

Using the binomial coefficient formula C(n, r) = n! / (r! * (n-r)!), we can simplify the terms with odd powers of x:

C(60, 1) * 1^59 * x^1 + C(60, 3) * 1^57 * x^3 + C(60, 5) * 1^55 * x^5 + ... + C(60, 59) * 1^1 * x^59

Now, notice that the coefficients of the terms with odd powers of x are symmetric. That is, the coefficient of x^1 is the same as the coefficient of x^59, the coefficient of x^3 is the same as the coefficient of x^57, and so on.

Therefore, we can pair up these terms and their coefficients and write the sum as:

(C(60, 1) * 1^59 + C(60, 3) * 1^57 + C(60, 5) * 1^55 + ... + C(60, 59) * 1^1) * (x^1 + x^3 + x^5 + ... + x^59)

Now, notice that the sum in the second part of the equation is actually the sum of an arithmetic series with a common difference of 2. So we can use the formula for the sum of an arithmetic series to simplify it:

Sum = (n/2) * (first term + last term) = (30/2) * (1 + 59) = 30 * 60 = 1800

Therefore, the sum of coefficients