Show that for each integer k ≥ 1, if P(k) is true, then P(k + 1) is true. Equations in Description below, fill in the right side.

Question

To show P(k) is true, f the left hand side of the equation is (1+(1/1))(1+(1/2))(1+(1/3))...(1+(1/k)), what is the right hand side equal to?

To show P(k+1) is true, if the left hand side of the equation is , what is the right hand side equal to?

Answer 1

I interpreted the question to be:

"Find a simplified expression in terms of k that is equal to (1+(1/1)) (1+(1/2)) (1+(1/3)) ... (1+(1/k)), and prove that the equality is true for any natural number k by Mathematical Induction."

First, let's calculate some of these for k = 1, 2, and 3.

When k=1, then (1 + (1/1)) = 2.

When k=2, then (1 + (1/1)) (1 + (1/2)) = 2 (3/2) = 3.

When k=3, then (1 + (1/1)) (1 + (1/2)) (1 + (1/3)) = 3 (4/3) = 4.

There seems to be the pattern that for any natural number k,

(1+(1/1)) (1+(1/2)) (1+(1/3)) ... (1+(1/k)) = k+1.

Let P(k) be the statement "(1+(1/1)) (1+(1/2)) (1+(1/3)) ... (1+(1/k)) = k+1".

Then we wish to prove that P(k) is true for any natural number k.

Let's prove this by Mathematical Induction.

We have already verified that P(1) is true ["When k=1, then (1 + (1/1)) = 2."]

Now we show the inductive step, that is: assuming that P(k) is true, show that P(k+1) is true.

(1+1/1) (1+1/2) (1+1/3) ... (1+1/k) (1+1/(k+1))

= [ (1+1/1) (1+1/2) (1+1/3) ... (1+1/k) ] (1+1/(k+1))

= (k+1) (1+ 1/(k+1)) [here, we have applied the inductive hypothesis that P(k) is true]

= (k+1) ( (k+1)/(k+1) + 1/(k+1) ) [writing the last factor as one fraction]

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

= k+2

Hence, P(k+1) will be true whenever P(k) is true.

Therefore, by Mathematical Induction, P(k) is true for any natural number k.