Question #390aa

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Answer 1 · 2016-05-09T16:23:13

#h_n = sqrt(n+1)#

We will proceed to show that #h_n=sqrt(n+1)# using induction.

First, for our base case, note that by the Pythagorean theorem, #h_1^2 = 1^1+1^1 = 2#, meaning #h_1 = sqrt(1+1)#, satisfying our equation.

Next, suppose that for some integer #k>=1# we have #h_k = sqrt(k+1)#. We must show that #h_(k+1) = sqrt((k+1)+1)#.

Indeed, as #h_(k+1)# is the hypotenuse of the right triangle with legs of lengths #1# and #h_k#, we have

#h_(k+1)^2 = 1^2 + h_k^2 = 1+sqrt(k+1)^2 = (k+1)+1#

Taking the square root of both sides, we obtain

#h_(k+1)=sqrt((k+1)+1)#

as desired. Thus, by induction, #h_n = sqrt(n+1)# for all #n>=1#.