expand ( 5d-1) 4

Do you mean (5d -1)^4 ?

Square 5d -1 twice.

The first squaring will give you
25d^2 -10d +1

Now square that. There will be more terms next time.

This is not trigonometry

Some people call the nth harmonic the nth ABOVE the fundamental and do not call the fundamental a harmonic. In that case, your answer would be 104/8 = 13 cm

Whoops. Ignore last answer. I posted it in the wrong place

To expand the expression (5d - 1)^4, we can use the binomial theorem. This theorem allows us to expand any binomial raised to a power. The binomial theorem states:

(x + y)^n = C(n,0) * x^(n - 0) * y^0 + C(n,1) * x^(n - 1) * y^1 + C(n,2) * x^(n - 2) * y^2 + ... + C(n,n) * x^(n - n) * y^n

Where C(n,k) represents the binomial coefficient, given by the formula:

C(n, k) = n! / (k! * (n - k)!)

Now, let's apply the binomial theorem to (5d - 1)^4:

(5d - 1)^4 = C(4,0) * (5d)^(4 - 0) * (-1)^0 + C(4,1) * (5d)^(4 - 1) * (-1)^1 + C(4,2) * (5d)^(4 - 2) * (-1)^2 + C(4,3) * (5d)^(4 - 3) * (-1)^3 + C(4,4) * (5d)^(4 - 4) * (-1)^4

Now let's calculate each term:

C(4,0) = 4! / (0! * (4 - 0)!) = 1
(5d)^(4 - 0) = (5d)^4 = 625d^4
(-1)^0 = 1

C(4,1) = 4! / (1! * (4 - 1)!) = 4
(5d)^(4 - 1) = (5d)^3 = 125d^3
(-1)^1 = -1

C(4,2) = 4! / (2! * (4 - 2)!) = 6
(5d)^(4 - 2) = (5d)^2 = 25d^2
(-1)^2 = 1

C(4,3) = 4! / (3! * (4 - 3)!) = 4
(5d)^(4 - 3) = (5d)^1 = 5d
(-1)^3 = -1

C(4,4) = 4! / (4! * (4 - 4)!) = 1
(5d)^(4 - 4) = (5d)^0 = 1
(-1)^4 = 1

Now we can substitute these values back into the binomial theorem expression:

(5d - 1)^4 = 1 * 625d^4 * 1 + 4 * 125d^3 * -1 + 6 * 25d^2 * 1 + 4 * 5d * -1 + 1 * 1 * 1

Simplifying this expression gives the expanded form:

(5d - 1)^4 = 625d^4 - 500d^3 + 150d^2 - 20d + 1