find the term containing y^3 in the expansion of (sqroot 2+y)^12
since the powers add up to 12, that would be the 10th term:
12C9 (√2)^9y^3 = 12C3 (16√2)y^3 = 3520√2 y^3
thank you
To find the term containing y^3 in the expansion of (sqrt(2) + y)^12, we can use the binomial theorem.
The binomial theorem states that for any non-negative integer n,
(x + y)^n = C(n,0) * x^n * y^0 + C(n,1) * x^(n-1) * y^1 + C(n,2) * x^(n-2) * y^2 + ... + C(n,n-1) * x^1 * y^(n-1) + C(n,n) * x^0 * y^n,
where C(n,k) is the binomial coefficient also known as "n choose k," given by C(n,k) = n! / (k!(n-k)!).
In this case, we are looking for the term containing y^3 in the expansion of (sqrt(2) + y)^12. So, we need to find the term with (sqrt(2))^k * y^(12-k) such that (12-k) = 3.
Substituting (12-k) = 3, we get k = 12 - 3 = 9.
Therefore, the term containing y^3 is C(12,9) * (sqrt(2))^9 * y^3.
Calculating C(12,9) = 12! / (9!(12-9)!) = 12! / (9! * 3!) = (12 * 11 * 10) / (3 * 2 * 1) = 220,
we can simplify the term to:
220 * (sqrt(2))^9 * y^3.
So, the term containing y^3 in the expansion of (sqrt(2) + y)^12 is 220 * 2^(9/2) * y^3.