Answer
for all integers n ≥ 1
(1 + 1/1) (1+1/2) (1+1/3) .....(1+1/n) = n + 1
Work Step by Step
let P(n) = all integers ≥ 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/n) = n + 1
Basis step: Show that P(1) is true:
P(1) = (1 + 1/1) = 2 by def
also P(1) = n + 1 = 1+1 = 2, so P(1) is true.
Inductive step: Show that for all integers k ≥ 1, if P(k) is true, then P(k+1) is true:
suppose P(k) = k+1is true (inductive hypothesis)
P(k+1) = (k+1) + 1 = k+2 (we need to get this result)
P(k+1) = 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/k) (1 + 1/(k+1)
but 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/k) = P(k) = k+1
so P(k+1) = (k+1) (1+1/(k+1)
= (k + 1) + (k+1)/(k+1) = k + 2 [which is what we need to show]
Therefore, P(n) is true for all integers n ≥ 1.