Find the third term of (2n+3m)^5
Same type of question as your previous two.
Let me know what you got
The 3rd term is
5C2 (2n)^(5-2) * (3m)^2 = 10*8n^3*9m^2 = 720n^3m^2
To find the third term of the expansion of (2n+3m)^5, we can use the binomial theorem. The binomial theorem states that the coefficient of the term with n^r * m^s is given by the formula:
C(r, s) * (2n)^(r-s) * (3m)^s
where C(r, s) is the binomial coefficient, also known as "n choose k," which can be calculated using the formula:
C(r, s) = r! / (s! * (r-s)!)
In this case, we want to find the third term, which corresponds to the case when r-s = 2. Let's calculate it step by step:
Step 1: Calculate C(r, s)
Using the formula, we have:
C(5, 2) = 5! / (2! * (5-2)!)
= 5! / (2! * 3!)
= (5 * 4 * 3!) / (2! * 3!)
= (5 * 4) / 2
= 10
Step 2: Calculate (2n)^(r-s)
Since r-s = 2, we have (2n)^(r-s) = (2n)^2 = 4n^2.
Step 3: Calculate (3m)^s
Since s = 2, we have (3m)^s = (3m)^2 = 9m^2.
Step 4: Multiply all the calculated terms
Now, multiply the results of Step 1, Step 2, and Step 3 together:
10 * 4n^2 * 9m^2 = 360n^2m^2
Therefore, the third term of the expansion of (2n+3m)^5 is 360n^2m^2.