write down the binomia expansion of (1+y)^8 where y=x+x² as far as the term x^6
(1+y)^8 , for y = x+x^2
= 1 + 8y + 28y^2 + 56y^3 + 70y^4 + 56y^5 + 70y^6 + ..
= 1 + 8(x+x^2) + 28(x+x^2)^2 + 56(x+x^2)^3 + 70(x+x^2)^6 + ...
( the remaining terms will contain x^7 and higher )
= take over, take care in the expansions
or
look up trinomial expansion here:
https://www.qc.edu.hk/math/Advanced%20Level/Trinomials.htm
great! i am very happy with this.... i will make sure i study it well
To expand the binomial (1+y)^8, where y = x + x^2, we can use the binomial theorem. Keep in mind that this expansion will include terms of increasing powers of y.
The binomial theorem states that for any positive integer n:
(1+y)^n = C(n,0)*1^n*y^0 + C(n,1)*1^(n-1)*y^1 + C(n,2)*1^(n-2)*y^2 + ... + C(n,k)*1^(n-k)*y^k + ... + C(n,n)*1^0*y^n
where C(n,k) represents the binomial coefficient, which can be calculated as:
C(n,k) = n! / (k! * (n-k)!)
In our case, we want to expand (1+y)^8, where y = x + x^2. To do this, substitute y = x + x^2 into the binomial theorem formula.
(1+y)^8 = C(8,0)*1^8*(x + x^2)^0 + C(8,1)*1^7*(x + x^2)^1 + C(8,2)*1^6*(x + x^2)^2 + ...
We can now start expanding the terms.
First, let's calculate the binomial coefficients:
C(8,0) = 8! / (0! * (8-0)!) = 1
C(8,1) = 8! / (1! * (8-1)!) = 8
C(8,2) = 8! / (2! * (8-2)!) = 28
Now, let's expand the first few terms:
Term 1: C(8,0)*1^8*(x + x^2)^0 = 1*(x + x^2)^0 = 1
Term 2: C(8,1)*1^7*(x + x^2)^1 = 8*(x + x^2)^1 = 8*(x + x^2)
Term 3: C(8,2)*1^6*(x + x^2)^2 = 28*(x + x^2)^2
To expand this further, we use the binomial theorem for (x + x^2)^2:
(x + x^2)^2 = (x^2 + 2*x*x^2 + x^4) = x^2 + 2x^3 + x^4
Expanding Term 3:
Term 3 = 28*(x + x^2)^2 = 28*(x^2 + 2x^3 + x^4) = 28x^2 + 56x^3 + 28x^4
So, the binomial expansion of (1+y)^8, where y = x + x^2, expanded as far as the term x^6, is:
(1+y)^8 = 1 + 8*(x + x^2) + 28x^2 + 56x^3 + 28x^4 + ... (terms beyond x^4)