a) Find a formula for 1/(1×2) + 1/(2×3) + 1/n(n+1) by examining the values of this expression for small values of n. b)Prove the formula you conjectured in part (a)

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Let’s solve this exercise!

a) Find a formula by examining the values of this expression for small values of n

n=0 it is undefined.

n=1: 1/(1×2) = 1/2

n=2: 1/(1×2) + 1/(2×3) = 1/2 + 1/6 = 4/6=2/3

n=3: 1/2 + 1/6 +1/12 = 3/4

n=4: 3/4 +1/20 = 4/5

1/(1×2) + 1/(2×3) + 1/n(n+1) = n/(n+1), for n>0

b)Prove the formula you conjectured in part (a)

To prove the formula above we are going to use mathematical induction. The reason is that we need to prove a formula (P(n)) is true for all positive numbers.

PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive integers k.” Source: Discrete Mathematics and its Applications by Rosen.

Basis step:

P(1): 1/(1×2) = 1/(1+1) = 1/2

Inductive hypothesis:

P(k): 1/(1×2) + 1/(2×3) + 1/k(k+1) = k/(k+1)

Inductive step:

P(k+1): 1/(1×2) + 1/(2×3) + 1/k(k+1) + 1/[(k+1)(k+2)] = (k+1)/(k+2)

1/(1×2) + 1/(2×3) + 1/k(k+1) + 1/[(k+1)(k+2)] applying the inductive hypothesis,

= k/(k+1) + 1/[(k+1)(k+2)], common factor 1/(k+1)

= [1/(k+1)] [k+1/(k+2)]

= [1/(k+1)] [k(k+2)+1]/(k+2), k(k+2)+1 = k2+2k+1 = (k+1)2, it follows

=[1/(k+1)] [(k+1)2]/(k+2), simplifying using (k+1)

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

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