1A> What is the coefficient of x^(2r) in the expansion of (1-x^2)^n, where n is a positive integer ?
1B> Find the coefficient of (1+x)^7 (1-x) ^8
Thanks for your help :)
1A> There is no x^(2r) term. Do you mean the x^2 term coefficient?
1B> Which coefficient do you want? You only ask for one, but there are 15 of them.
This web site will tell you how to use the binomial theorem to get the coefficients of polynomials:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut54_bi_theor.htm
in the first part, r is any integer.
in the second part, i wanna find the coefficient of x^7
actually my teacher didn't teach me anything about binomial theorem. This topic is self-studied.
We did this in Algebra III, but I can not remember how to do it. What course does this fall under for you, Tommy? Is there a section in your textbook that mentions this, possibly as an extension or application or some such extra bonus material?
It is additional maths.
I learnt the binomial theorem, the (r+1) term theorem, how to expand the binomial expression using the the pascal triangle.
pls help me with this binomial expansion
1A> To find the coefficient of x^(2r) in the expansion of (1-x^2)^n, we can use the binomial theorem. According to the binomial theorem, the coefficient of x^r in the expansion of (a + b)^n is given by the formula: C(n, r) * a^(n-r) * b^r, where C(n, r) represents the binomial coefficient.
In this case, we have (1 - x^2)^n, so a = 1 and b = -x^2. We want to find the coefficient of x^(2r), so r = 2r. Using the formula above, the coefficient of x^(2r) would be C(n, 2r) * 1^(n-2r) * (-x^2)^2r.
Therefore, the coefficient of x^(2r) in the expansion of (1 - x^2)^n is C(n, 2r) * (-1)^(2r) * x^(4r).
1B> To find the coefficient of (1 + x)^7 (1 - x)^8, we can expand the expression using the binomial theorem. According to the binomial theorem, the expansion of (a + b)^n is given by the formula: C(n, 0) * a^n + C(n, 1) * a^(n-1) * b + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n) * b^n.
In this case, we have (1 + x)^7 (1 - x)^8, which can be written as (1 + x)^7 * (1 - x)^8. We can expand each binomial separately using the formula above.
Expanding (1 + x)^7, we have:
C(7, 0) * 1^7 + C(7, 1) * 1^6 * x + C(7, 2) * 1^5 * x^2 + C(7, 3) * 1^4 * x^3 + C(7, 4) * 1^3 * x^4 + C(7, 5) * 1^2 * x^5 + C(7, 6) * 1^1 * x^6 + C(7, 7) * x^7.
Expanding (1 - x)^8, we have:
C(8, 0) * 1^8 + C(8, 1) * 1^7 * (-x) + C(8, 2) * 1^6 * (-x)^2 + C(8, 3) * 1^5 * (-x)^3 + C(8, 4) * 1^4 * (-x)^4 + C(8, 5) * 1^3 * (-x)^5 + C(8, 6) * 1^2 * (-x)^6 + C(8, 7) * 1^1 * (-x)^7 + C(8, 8) * (-x)^8.
Next, we multiply the two expansions by substituting the corresponding values of a and b:
(1^7 + 7*1^6 * x + 21*1^5 * x^2 + 35*1^4 * x^3 + 35*1^3 * x^4 + 21*1^2 * x^5 + 7*1^1 * x^6 + 1*x^7) * (1^8 - 8*1^7 * x + 28*1^6 * x^2 - 56*1^5 * x^3 + 70*1^4 * x^4 - 56*1^3 * x^5 + 28*1^2 * x^6 - 8*1^1 * x^7 + 1*(-x)^8).
Finally, we multiply each term and combine like terms to simplify the expression and find the coefficient of the desired term, which in this case is the coefficient of x^15.