The polynomial function defined by

f(x) = 1 − x + x^2 − x^3 + ... + x^16 − x^17

may be written in the form

f(y) = a_0 + a_1 y + a_2 y^2 + ... + a_16 y^16 + a_17 y^17,

where y = x + 1 and a_i s are constants. Find a_2.

To find the constant term a₂ in the polynomial function f(y), we need to substitute y = x + 1 into the given polynomial function f(x):

f(y) = 1 − (x + 1) + (x + 1)² − (x + 1)³ + ... + (x + 1)¹⁶ − (x + 1)¹⁷

The goal is to determine the coefficient of y², which corresponds to the term a₂y² in f(y). So, we need to look for the term that contains y².

Expanding each term of the polynomial above, we get:

f(y) = 1 − x - 1 + x² + 2x + 1 − x³ - 3x² - 3x - x² - 1 + 3x + 3x² + ... + (x¹⁶ + 16x¹⁵ + 120x¹⁴ + ...)

Now, we can simplify the expression by combining like terms:

f(y) = 1 - x³ + (1 - 1) + (2x + 3x + 4x + ... + 16x¹⁵) + (1 - 1) + (3x + 6x + ... + 120x¹⁴) + ...

Notice that in the expression (2x + 3x + 4x + ... + 16x¹⁵), we have terms with coefficients that form an arithmetic series. Using the formula for the sum of an arithmetic series, we can find the sum of these terms:

Sum = n/2 * (first term + last term)
= 15/2 * (2x + 16x¹⁵)
= 15 * (x + 8x¹⁵)

Continuing with the simplification of f(y), we have:

f(y) = 1 - x³ + 0 + 15 * (x + 8x¹⁵) + 0 + ...

Now, we can rewrite f(y) in the desired form:

f(y) = a₀ + a₁y + a₂y² + ... + a₁₆y¹⁶ + a₁₇y¹⁷

Comparing coefficients of the terms on both sides, we can see that:

a₀ = 1
a₁ = -1
a₂ = 0
a₃ = -1
a₄ = 0
...

Therefore, the constant term a₂ is 0.