1) Use the Binomial Theorem to write out the expansion of (a+b+c)^4. Hint think of (a+b+c)^4 as [(a+b]) +c]^4.
2) Use the Binomial Theorem to write out the expansion of (x−2+y)^3. Think of (x−2+y)^3 as [(x−2)+y]^3
I'll get you started on the 2nd one.
[(x−2)+y]^3
= (x-2)^3 + 3(x-2)^2 y + 3(x-2) y^2 + y^3
= ...
The first one is even easier.
1) To expand (a+b+c)^4 using the Binomial Theorem, we can consider [(a+b)+c]^4.
The Binomial Theorem states that for any positive integer n, the expansion of (x+y)^n is given by:
(x+y)^n = C(n, 0)x^n + C(n, 1)x^(n-1)y + C(n, 2)x^(n-2)y^2 + ... + C(n, n)y^n
where C(n, k) represents the binomial coefficient, which is equal to n! / (k!(n-k)!)
Using this formula, we can expand [(a+b)+c]^4:
[(a+b)+c]^4 = C(4, 0)(a+b)^4c^0 + C(4, 1)(a+b)^3c^1 + C(4, 2)(a+b)^2c^2 + C(4, 3)(a+b)^1c^3 + C(4, 4)(a+b)^0c^4
Simplifying, we get:
(a+b+c)^4 = C(4, 0)(a^4) + C(4, 1)(a^3)(b) + C(4, 2)(a^2)(b^2) + C(4, 3)(a)(b^3) + C(4, 4)(b^4) + C(4, 0)(a^4)(c) + C(4, 1)(a^3)(b)(c) + C(4, 2)(a^2)(b^2)(c) + C(4, 3)(a)(b^3)(c) + C(4, 4)(b^4)(c)
2) To expand (x-2+y)^3 using the Binomial Theorem, we can consider [(x-2)+y]^3.
Using the same formula as before:
[(x-2)+y]^3 = C(3, 0)(x-2)^3y^0 + C(3, 1)(x-2)^2y^1 + C(3, 2)(x-2)^1y^2 + C(3, 3)(x-2)^0y^3
Simplifying further:
(x-2+y)^3 = C(3, 0)(x^3) + C(3, 1)(x^2)(-2) + C(3, 2)(x)(-2)^2 + C(3, 3)(-2)^3 + C(3, 0)(x^3)(y) + C(3, 1)(x^2)(-2)(y) + C(3, 2)(x)(-2)^2(y) + C(3, 3)(-2)^3(y)
1) To expand (a+b+c)^4 using the Binomial Theorem, we need to find the coefficients of each term in the expansion.
The Binomial Theorem states that the expansion of (a+b)^n can be written as the sum of the binomial coefficients multiplied by the powers of a and b, where n is a positive integer.
Given (a+b+c)^4, we can rewrite it as [(a+b)+c]^4 to make it easier to apply the theorem. Now we can see that we have two terms within the bracket: (a+b) and c.
Now, we can write out the expansion using the Binomial Theorem:
(a+b+c)^4 = [(a+b)+c]^4 = C(4,0)((a+b)^4)(c^0) + C(4,1)((a+b)^3)(c^1) + C(4,2)((a+b)^2)(c^2) + C(4,3)((a+b)^1)(c^3) + C(4,4)((a+b)^0)(c^4)
Let's break it down further:
- C(4,0) represents the binomial coefficient for selecting 0 terms from (a+b) and 4 terms from c. This coefficient is 1.
- C(4,1) represents the binomial coefficient for selecting 1 term from (a+b) and 3 terms from c. This coefficient is 4.
- C(4,2) represents the binomial coefficient for selecting 2 terms from (a+b) and 2 terms from c. This coefficient is 6.
- C(4,3) represents the binomial coefficient for selecting 3 terms from (a+b) and 1 term from c. This coefficient is 4.
- C(4,4) represents the binomial coefficient for selecting 4 terms from (a+b) and 0 terms from c. This coefficient is 1.
Now, we substitute the respective coefficients and simplify the expression:
(a+b+c)^4 = 1((a+b)^4)(c^0) + 4((a+b)^3)(c^1) + 6((a+b)^2)(c^2) + 4((a+b)^1)(c^3) + 1((a+b)^0)(c^4)
Expand each of the terms according to the powers of (a+b):
(a+b+c)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) + c(4a^3 + 12a^2b + 12ab^2 + 4b^3) + c^2(6a^2 + 12ab + 6b^2) + c^3(4a + 4b) + c^4
This is the expanded form of (a+b+c)^4 using the Binomial Theorem.
2) To expand (x−2+y)^3 using the Binomial Theorem, we can follow a similar approach.
Think of (x−2+y)^3 as [(x−2)+y]^3. Now we have two terms within the bracket: (x−2) and y.
Using the Binomial Theorem, we can write out the expansion:
(x−2+y)^3 = [(x−2)+y]^3 = C(3,0)((x−2)^3)(y^0) + C(3,1)((x−2)^2)(y^1) + C(3,2)((x−2)^1)(y^2) + C(3,3)((x−2)^0)(y^3)
Let's break it down further:
- C(3,0) represents the binomial coefficient for selecting 0 terms from (x−2) and 3 terms from y. This coefficient is 1.
- C(3,1) represents the binomial coefficient for selecting 1 term from (x−2) and 2 terms from y. This coefficient is 3.
- C(3,2) represents the binomial coefficient for selecting 2 terms from (x−2) and 1 term from y. This coefficient is 3.
- C(3,3) represents the binomial coefficient for selecting 3 terms from (x−2) and 0 terms from y. This coefficient is 1.
Now, substitute the respective coefficients and simplify the expression:
(x−2+y)^3 = 1((x−2)^3)(y^0) + 3((x−2)^2)(y^1) + 3((x−2)^1)(y^2) + 1((x−2)^0)(y^3)
Expand each of the terms according to the powers of (x−2):
(x−2+y)^3 = (x^3 - 6x^2 + 12x - 8) + 3(y)(x^2 - 4x + 4) + 3(y^2)(x - 2) + y^3
This is the expanded form of (x−2+y)^3 using the Binomial Theorem.