Expand (1-6x)^4 (1+2x)^7 in ascending powers of x up to and includeing the terms in x^3.
(1-6x)^4
= 1^4 + 4*1^3(-6x)^1 + 6*1^2(-6x)^2 + 4(1^1(-6x)^3 + 1(-6x)^4
= 1-24x+216x^2-864x^3+...
(1+2x)^7 = 1+14x+84x^2+280x^3+...
Now just work from the left
1(1+14x+84x^2+280x^3)
-24x(1+14x+84x^2+280x^3)
+216x^2(1+14x+84x^2+280x^3)
-864x^3(1+14x+84x^2+280x^3)
...
1-10x-36x^2+424x^3+...
To expand the expression (1-6x)^4 (1+2x)^7 in ascending powers of x up to and including the terms in x^3, we can use the binomial theorem.
The binomial theorem states that (a + b)^n can be expanded using the following formula:
(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!/((n-k)!*k!).
Let's expand (1-6x)^4 and (1+2x)^7 using this formula:
(1-6x)^4 = C(4, 0)*1^4*(-6x)^0 + C(4, 1)*1^3*(-6x)^1 + C(4, 2)*1^2*(-6x)^2 + C(4, 3)*1^1*(-6x)^3 + C(4, 4)*1^0*(-6x)^4
(1+2x)^7 = C(7, 0)*1^7*(2x)^0 + C(7, 1)*1^6*(2x)^1 + C(7, 2)*1^5*(2x)^2 + C(7, 3)*1^4*(2x)^3 + C(7, 4)*1^3*(2x)^4 + C(7, 5)*1^2*(2x)^5 + C(7, 6)*1^1*(2x)^6 + C(7, 7)*1^0*(2x)^7
Let's calculate the individual terms:
(1-6x)^4 = 1 - 24x + 216x^2 - 864x^3 + 1296x^4
(1+2x)^7 = 1 + 14x + 98x^2 + 392x^3 + 980x^4 + 1568x^5 + 1792x^6 + 1280x^7
Now, let's multiply the two expressions term by term:
(1-6x)^4 (1+2x)^7 = (1 - 24x + 216x^2 - 864x^3 + 1296x^4) * (1 + 14x + 98x^2 + 392x^3 + 980x^4 + 1568x^5 + 1792x^6 + 1280x^7)
Expanding this expression will involve multiplying each term from the first expansion by every term from the second expansion. However, since we only need the terms up to and including x^3, we will only keep those terms:
(1-6x)^4 (1+2x)^7 = 1 + (14 - 24)x + (98 - 168 + 216)x^2 + (392 - 672 + 864 - 504)x^3
Simplifying further:
(1-6x)^4 (1+2x)^7 = 1 - 10x + 146x^2 - 920x^3
Therefore, the expanded expression in ascending powers of x up to and including the terms in x^3 is:
(1-6x)^4 (1+2x)^7 = 1 - 10x + 146x^2 - 920x^3
To expand the expression (1-6x)^4 (1+2x)^7 and obtain the terms up to and including x^3, we will use the binomial theorem. The binomial theorem states that for any real numbers a and b, and any non-negative 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 and can be calculated as:
C(n, k) = n! / (k!(n-k)!)
Let's apply the binomial theorem to expand (1-6x)^4 (1+2x)^7:
First, we expand (1-6x)^4:
(1-6x)^4 = C(4, 0)(1^4)(-6x)^0 + C(4, 1)(1^3)(-6x)^1 + C(4, 2)(1^2)(-6x)^2 + C(4, 3)(1^1)(-6x)^3 + C(4, 4)(1^0)(-6x)^4
= (1)(1)(1) + (4)(1)(-6x) + (6)(1)(36x^2) + (4)(1)(-216x^3) + (1)(1)(1296x^4)
Now, we expand (1+2x)^7:
(1+2x)^7 = C(7, 0)(1^7)(2x)^0 + C(7, 1)(1^6)(2x)^1 + C(7, 2)(1^5)(2x)^2 + C(7, 3)(1^4)(2x)^3 + C(7, 4)(1^3)(2x)^4 + C(7, 5)(1^2)(2x)^5 + C(7, 6)(1^1)(2x)^6 + C(7, 7)(1^0)(2x)^7
= (1)(1)(1) + (7)(1)(2x) + (21)(1)(4x^2) + (35)(1)(8x^3) + (35)(1)(16x^4) + (21)(1)(32x^5) + (7)(1)(64x^6) + (1)(1)(128x^7)
Now, we multiply these two expanded expressions together, term by term, and combine like terms. To find the terms up to and including x^3, we will keep only those terms that have x^3 or lower powers:
(1-6x)^4 (1+2x)^7 = [1 + (-24x) + 216x^2 + (-864x^3)] + [7(2x) + (-168x^2) + 1344x^3] + [21(4x^2) + (-504x^3)] + [35(8x^3)] + 0x^4 + 0x^5 + 0x^6 + 0x^7
Simplifying further, we have:
= 1 - 24x + 216x^2 - 864x^3 + 14x + 168x^2 - 1344x^3 + 84x^2 - 2016x^3 + 280x^3
= 1 - 24x + 14x + 216x^2 + 168x^2 + 84x^2 - 864x^3 - 1344x^3 - 2016x^3 + 280x^3
= 1 - 10x + 468x^2 - 3644x^3
Thus, the expansion of (1-6x)^4 (1+2x)^7 in ascending powers of x up to and including the terms in x^3 is: 1 - 10x + 468x^2 - 3644x^3.