Given the following finite sum

1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+....+1/(n(n+1))

a) Find the first 5 partial sums
b) Make a conjecture for a formula for the sum of the first n terms
c) Use mathematical induction to prove your formula

a) To find the first 5 partial sums, we can simply substitute different values for n in the given expression and calculate the sums.

For n = 1:
S₁ = 1/(1*2) = 1/2

For n = 2:
S₂ = 1/(1*2) + 1/(2*3) = 1/2 + 1/6 = 2/3

For n = 3:
S₃ = 1/(1*2) + 1/(2*3) + 1/(3*4) = 1/2 + 1/6 + 1/12 = 3/4

For n = 4:
S₄ = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) = 1/2 + 1/6 + 1/12 + 1/20 = 4/5

For n = 5:
S₅ = 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 = 5/6

b) From the given expression, we can observe that each term is of the form 1/(n(n+1)). We can rewrite this as (n+1 - n)/(n(n+1)) = 1/n - 1/(n+1). Therefore, the terms form a telescoping series, where all the intermediate terms cancel out. Thus, we can rewrite the sum of the first n terms as:

Sₙ = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

Simplifying this expression, we can see that most of the terms cancel out, leaving only the first term (1/1) and the last term (-1/(n+1)):

Sₙ = 1 - 1/(n+1) = (n+1)/(n+1) - 1/(n+1) = n/(n+1)

Therefore, our conjecture for a formula for the sum of the first n terms is:

Sₙ = n/(n+1)

c) To prove our formula using mathematical induction, we need to show that it holds true for the base case (n = 1) and assume it is true for some value k, then prove it for k + 1.

Base case (n = 1):
S₁ = 1/(1+1) = 1/2
The formula S₁ = 1/(1+1) = 1/2 holds true for n = 1.

Assuming the formula is true for k:
Sₖ = k/(k+1)

Now, let's prove it for k + 1:
Sₖ₊₁ = (k+1)/(k+1+1) = (k+1)/(k+2)

We need to show that Sₖ₊₁ = (k+1)/(k+2) equals Sₖ + 1/(k+1).

Sₖ + 1/(k+1) = k/(k+1) + 1/(k+1)
= (k+1)/(k+1)
= (k+1)/(k+2)

Since Sₖ + 1/(k+1) = (k+1)/(k+2) = Sₖ₊₁, the formula holds true for k + 1.

By mathematical induction, the formula Sₙ = n/(n+1) holds for all natural numbers n.