Answer:
The expansion of (a + b)^6 is a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + 6ab^5 + b^6.
Option (a) is true.
Explanation:
The binomial theorem states that for any positive integer n:
(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, given by:
C(n, k) = n! / (k!(n-k)!)
In this case, we want to expand (a + b)^6, so n = 6.
(a + b)^6 = C(6, 0)a^6 b^0 + C(6, 1)a^5 b^1 + C(6, 2)a^4 b^2 + C(6, 3)a^3 b^3 + C(6, 4)a^2 b^4 + C(6, 5)a^1 b^5 + C(6, 6)a^0 b^6
Expanding each term using the binomial coefficient formula:
(a + b)^6 = (1)a^6 (1) + (6)a^5 b^1 + (15)a^4 b^2 + (20)a^3 b^3 + (15)a^2 b^4 + (6)a^1 b^5 + (1)a^0 b^6
Simplifying each term:
(a + b)^6 = a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + 6ab^5 + b^6
Therefore,
Option (a) is true.
Question: Use the binomial theorem to expand (a + b)^6.
a) a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + 6ab^5 + b^6
b) a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + 6ab^5
c) a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 6ab^5 + b^6
d) a^6 + 6a^5 b + 15a^4 b^2 + 20a^3 b^3 + 15a^2 b^4 + b^6