expand [e+1/e]^4 in terms of e.
express [e+1/e]^4 + [e-1/e]^4 as the sum of three terms.
please help
(e + 1/e)^4
= e^4 + 4e^3/e + 6e^2/e^2 + 4e/e^3 + 1/e^4
= e^4 + 4e^2 + 6 + 4/e^2 + 1/e^4
for the 2nd one, since the terms alternate in sign, every other term cancels out, leaving just
2(e^4 + 6 + 1/e^4)
Sure! Let's start by expanding [e+1/e]^4 in terms of e:
[e+1/e]^4 = e^4 + 4e^3(1/e) + 6e^2(1/e)^2 + 4e(1/e)^3 + (1/e)^4
= e^4 + 4e + 6 + 4e^-1 + e^-4
Now, let's express [e+1/e]^4 + [e-1/e]^4 as the sum of three terms:
[e+1/e]^4 + [e-1/e]^4 = (e^4 + 4e + 6 + 4e^-1 + e^-4) + (e^4 - 4e + 6 - 4e^-1 + e^-4)
= 2e^4 + 12 + 2e^-4
So, [e+1/e]^4 + [e-1/e]^4 can be expressed as the sum of three terms: 2e^4, 12, and 2e^-4.
To expand the expression [e+1/e]^4 in terms of e, we can use the binomial theorem.
The binomial theorem states that for any real number a and a positive integer n, we have (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, a is e, and b is 1/e. So we have:
[e+1/e]^4 = C(4,0)*e^4*(1/e)^0 + C(4,1)*e^3*(1/e)^1 + C(4,2)*e^2*(1/e)^2 + C(4,3)*e^1*(1/e)^3 + C(4,4)*e^0*(1/e)^4
Simplifying each term:
[e+1/e]^4 = (1)*e^4 + (4)*e^3*(1/e) + (6)*e^2*(1/e)^2 + (4)*e^1*(1/e)^3 + (1)*e^0*(1/e)^4
[e+1/e]^4 = e^4 + 4e^2 + 6 + 4e^(-2) + e^(-4)
To express [e+1/e]^4 + [e-1/e]^4 as the sum of three terms, we can group like terms together. From the above expansion, the terms involving e can be grouped as follows:
[e+1/e]^4 + [e-1/e]^4 = (e^4 + e^4) + (4e^2 + 4e^(-2)) + (6)
= 2e^4 + 4e^2 + 4e^(-2) + 6
So, [e+1/e]^4 + [e-1/e]^4 can be expressed as the sum of three terms:
[e+1/e]^4 + [e-1/e]^4 = 2e^4 + 4e^2 + 4e^(-2) + 6
To expand the expression [e+1/e]^4 in terms of e, we can use the binomial theorem. The binomial theorem states that for any real number a and 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 our case, a = e and b = 1/e. Plugging these values into the binomial theorem, we have:
[e+1/e]^4 = C(4,0) * e^4 * (1/e)^0 + C(4,1) * e^3 * (1/e)^1 + C(4,2) * e^2 * (1/e)^2 + C(4,3) * e^1 * (1/e)^3 + C(4,4) * e^0 * (1/e)^4.
Now, let's simplify each term:
C(4,0) = 1,
C(4,1) = 4,
C(4,2) = 6,
C(4,3) = 4,
C(4,4) = 1.
And we have:
[e+1/e]^4 = e^4 + 4e^3 * 1/e + 6e^2 * 1/e^2 + 4e * 1/e^3 + 1 * 1/e^4.
Now, simplify each term:
e^4 + 4e^3 * 1/e + 6e^2 * 1/e^2 + 4e * 1/e^3 + 1 * 1/e^4
= e^4 + 4e^2 + 6 + 4e + 1/e^2.
Therefore, [e+1/e]^4 expanded in terms of e is: e^4 + 4e^2 + 6 + 4e + 1/e^2.
To express [e+1/e]^4 + [e-1/e]^4 as the sum of three terms, we can simply assign new variables to each term and group them accordingly. Let's denote [e+1/e]^4 as A and [e-1/e]^4 as B.
We have:
A = e^4 + 4e^2 + 6 + 4e + 1/e^2
B = [e-1/e]^4.
Now, we can express the sum of the two terms as:
A + B = (e^4 + 4e^2 + 6 + 4e + 1/e^2) + [e-1/e]^4.
We can further simplify it by factoring out common terms:
A + B = e^4 + [e-1/e]^4 + 4e^2 + 4e + 6 + 1/e^2.
Now, we can regroup the terms:
A + B = (e^4 + [e-1/e]^4) + (4e^2 + 4e + 6) + (1/e^2).
Therefore, the sum of the two terms [e+1/e]^4 + [e-1/e]^4 can be expressed as the sum of three terms: (e^4 + [e-1/e]^4), (4e^2 + 4e + 6), and (1/e^2).