How do u expand this?
(a+b+c)^n
This should get you started.
http://en.wikipedia.org/wiki/Trinomial_expansion
If the ^ is a multiplication symbol, and you mean "expand" by using the distributive property, then this is the answer.
n(a+b+c) is the same as:
n(a) + n(b) + n(c)
Which is na+nb+nc
Hopefully, you know what this distributive property is, That would be VERY VERY helpful. I'll try to explain:
If you said 6(5-3), according to PEMDAS, you do the parenthesis first.
6(5-3)
6(2)
12
Okay? But, according to the distributive property, you can say that:
6(5-3) = 6(5) - 6(3)
So,
6(5) - 6(3)
30 - 18
12
See? Whether you use PEMDAS or the distributive property, you get the same answer.
To expand the expression (a+b+c)^n, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a+b)^n can be written as the sum of terms of the form (n choose k) * a^(n-k) * b^k, where (n choose k) is the binomial coefficient given by n! / (k! * (n-k)!), and n! represents the factorial of n.
In the case of (a+b+c)^n, we can consider it as (a+(b+c))^n and apply the binomial theorem. Let's break down the steps using an example with n = 3:
Step 1: Determine the binomial coefficients.
For n = 3, the binomial coefficients are (3 choose 0) = 1, (3 choose 1) = 3, (3 choose 2) = 3, and (3 choose 3) = 1.
Step 2: Write out the expansion.
The expansion of (a+b+c)^3 can be written as:
(a+b+c)^3 = (1*a^3*b^0*c^0) + (3*a^2*b^1*c^0) + (3*a^1*b^2*c^0) + (1*a^0*b^3*c^0) + (3*a^2*b^0*c^1) + (6*a^1*b^1*c^1) + (3*a^0*b^2*c^1) + (1*a^1*b^0*c^2) + (3*a^0*b^1*c^2) + (1*a^0*b^0*c^3)
Step 3: Simplify the terms.
Simplifying each term, we get:
(a+b+c)^3 = a^3 + 3*a^2*b + 3*a^2*c + 3*a*b^2 + 6*a*b*c + 3*a*c^2 + b^3 + 3*b^2*c + 3*b*c^2 + c^3
Therefore, the expanded form of (a+b+c)^n is:
(a+b+c)^n = a^n + (n choose 1) * a^(n-1) * b + (n choose 1) * a^(n-1) * c + (n choose 2) * a^(n-2) * b^2 + ... + (n choose i) * a^(n-i) * b^i * c^(n-i) + ... + (n choose n-1) * a * b^(n-1) * c + b^n + (n choose 1) * b^(n-1) * c + ... + (n choose n-1) * b * c^(n-1) + c^n