Find the term not involving y in the expansion of (1 + y2)^5
y^3
is your expression (1 + y^2)^5(y^3) ?
if so, then there is a y in every term of the expansion.
Check your typing.
To find the term not involving y in the expansion of (1 + y^2)^5, we need to expand the expression using the binomial theorem. The binomial theorem states that for any real numbers a and b, and any positive integer n:
(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-2) * a^2 * b^(n-2) + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) is the binomial coefficient, given by:
C(n, k) = n! / (k! * (n-k)!)
In this case, a = 1, b = y^2, and n = 5. We are interested in the term not involving y, which means the power of y should be 0.
Using the binomial theorem, we can rewrite the expansion as:
(1 + y^2)^5 = C(5, 0) * 1^5 * (y^2)^0 + C(5, 1) * 1^4 * (y^2)^1 + C(5, 2) * 1^3 * (y^2)^2 + C(5, 3) * 1^2 * (y^2)^3 + C(5, 4) * 1^1 * (y^2)^4 + C(5, 5) * 1^0 * (y^2)^5
Simplifying each term, we get:
(1 + y^2)^5 = 1 + 5y^2 + 10y^4 + 10y^6 + 5y^8 + y^10
The term not involving y is the constant term, which is 1 in this case.