Solution:

We can expand the given expression using binomial theorem as follows:

(1 + x + x² + x³)⁵

= (1 + x + x² + x³)(1 + x + x² + x³)(1 + x + x² + x³)(1 + x + x² + x³)(1 + x + x² + x³)

Using the distributive property, we can expand this expression into a sum of terms, each of which is a product of some power of 1, x, x², and x³. The coefficient of each term is the product of the coefficients of the factors that make up the term.

To find the sum of the coefficients of even powers of x, we need to add up the coefficients of the terms that have an even power of x. In other words, we need to add up the coefficients of the terms of the form x²ⁿ, where n is a non-negative integer.

We can simplify this task by noticing that, for each power of x, there are only four terms that contribute to the coefficient of that power. For example, the coefficient of x² is the sum of the coefficients of the terms 1x²x²x², x²1x²x², x²x²1x², and 1x²1x²x².

Therefore, we can find the sum of the coefficients of the terms of the form x²ⁿ by summing the coefficients of the terms of the form 1x²x²x², x²1x²x², x²x²1x², and 1x²1x²x², and then multiplying the result by 4.

Let's compute this sum. The coefficient of the term 1x²x²x² is 1*1*1*1*1=1. Similarly, the coefficient of each of the other three terms is also 1. Therefore, the sum of the coefficients of the terms of the form x²ⁿ is 4.

To find the sum of the coefficients of even powers of x, we need to add up