Find the third term of (2n+3m)^5
720n^3m^2 I received as my answer.
and you have a problem with it?
Does it fit with what you know about binomial expansions?
You were correct yesterday and are still correct today.
To find the third term of the expression (2n + 3m)^5, we can use the binomial theorem. The binomial theorem states that for any two numbers a and b and a positive integer n, the expansion of (a + b)^n can be written as:
(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-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) represents the binomial coefficient, which is calculated as C(n, k) = n! / (k! * (n-k)!).
In this case, we need to find the third term, so we are interested in the term with a^(5-2) * b^2, or in other words, a^3 * b^2. In the expression (2n + 3m)^5, a is 2n and b is 3m.
Using the binomial coefficients, we can calculate the third term as:
C(5, 2) * (2n)^3 * (3m)^2
C(5, 2) = 5! / (2! * (5-2)!) = 10
Substituting the values:
10 * (2n)^3 * (3m)^2
Simplifying further:
10 * 8n^3 * 9m^2
We can now simplify the expression:
10 * 72n^3m^2 = 720n^3m^2
So the third term of the expansion (2n + 3m)^5 is 720n^3m^2.