Question

(a-2b)^5 решите по формуле бинома

Answer 1

$(a-2b)^5=a^5-10a^4b+40a^3b^2-80a^2b^3+80ab^4-32b^5$

Answer 2

To expand the expression (a-2b)^5 using the binomial theorem, we can use the following formula:

The binomial theorem states that for any real numbers a and b, and any positive integer n, the expansion of (a + b)^n is given by:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k

where Σ denotes the summation, (n choose k) is the binomial coefficient (n! / (k! * (n-k)!)), and the summation runs from k = 0 to k = n.

Let's apply this formula to our expression (a-2b)^5:

n = 5 a = a b = -2b

The expansion will be:

(a - 2b)^5 = Σ (5 choose k) * a^(5-k) * (-2b)^k, for k = 0 to k = 5

Let's calculate the terms for each value of k and then sum them up:

k = 0: (5 choose 0) * a^5 * (-2b)^0 = 1 * a^5 * 1 = a^5

k = 1: (5 choose 1) * a^4 * (-2b)^1 = 5 * a^4 * (-2b) = -10a^4b

k = 2: (5 choose 2) * a^3 * (-2b)^2 = 10 * a^3 * (4b^2) = 40a^3b^2

k = 3: (5 choose 3) * a^2 * (-2b)^3 = 10 * a^2 * (-8b^3) = -80a^2b^3

k = 4: (5 choose 4) * a^1 * (-2b)^4 = 5 * a * (16b^4) = 80ab^4

k = 5: (5 choose 5) * a^0 * (-2b)^5 = 1 * 1 * (-32b^5) = -32b^5

Now, let's put all the terms together:

(a - 2b)^5 = a^5 - 10a^4b + 40a^3b^2 - 80a^2b^3 + 80ab^4 - 32b^5

So, the expanded form of (a-2b)^5 using the binomial theorem is:

a^5 - 10a^4b + 40a^3b^2 - 80a^2b^3 + 80ab^4 - 32b^5