00:01
Problem we want to prove this statement using mathematical induction the term a n can be found by using the recursive statement a n minus one the previous term plus one over n times n plus one and it can also be found by using the expression n over n plus one so the base case will be for n equals two because it says n greater than one uh for n equals two the left side would be a one plus one over two times three because n is two.
00:32
A1 is given to us, that's one half plus one over six.
00:39
So that's three over six plus one over six, so four over six, which is the same as two over three.
00:46
On the other side of the equation, if n is two, then that would be two over two plus one or three.
00:51
So since both are equal, we know that a1 is true.
00:59
Next, the inductive step, we assume a k is true and if a k is true that means a k is defined by a k minus 1 plus 1 over k times k plus 1 equals k over k plus 1 then for a k plus 1 using the recursive formula that would be equal to a k plus 1 over k plus 1 and k plus 2.
01:42
Ak, we've found an equivalent expression for that...