FInd the 5th term of (x2y)^10

There is only one term.

It is x^20*y^10

Did it really ask for the fifth term of (x^2 + y)^10 ?

That would depend upon the order that the terms are listed

Taking a look at the previous posting plug your values into row 10 of Pascal's triangle:

1 10 45 120 210 252 210 120 45 10 1

Watch out, because the terms involve x^2 and not just x:

(x^2)^10 + 10(x^2)^9 y + ...

The 5th term is 210(x^2)^6y^4 = 210 x^12 y^4

To find the 5th term of the expression (x^2y)^10, we can use the binomial theorem. The binomial theorem states that for any expression of the form (a + b)^n, the expansion can be calculated using the formula:

(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, r) * a^(n-r) * b^r + ... + C(n, n) * a^0 * b^n

Where C(n, r) represents the binomial coefficient, and is equal to n! / (r! * (n - r)!)

In this case, the expression can be rewritten as:

(x^2y)^10 = (x^2)^10 * y^10 = x^20 * y^10

To find the 5th term, we need to find the term that corresponds to r = 4. Using the binomial theorem formula, the 5th term can be calculated as:

C(10, 4) * x^(20-4) * y^4

Let's calculate it step by step:

Calculating the binomial coefficient:
C(10, 4) = 10! / (4! * (10 - 4)!) = 10! / (4! * 6!) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210

Substituting the values into the formula:
5th term = 210 * x^(20-4) * y^4
= 210 * x^16 * y^4

Therefore, the 5th term of the expression (x^2y)^10 is 210 * x^16 * y^4.