SOLUTION: Prove by mathematical induction {{{1/(1*2) + 1/(2*3) + 1/(3*4)}}}+ ...+{{{1/n(n+1) = n/(n+1)}}}

Algebra ->  Sequences-and-series -> SOLUTION: Prove by mathematical induction {{{1/(1*2) + 1/(2*3) + 1/(3*4)}}}+ ...+{{{1/n(n+1) = n/(n+1)}}}       Log On


   

Question 862408: Prove by mathematical induction
1%2F%281%2A2%29+%2B+1%2F%282%2A3%29+%2B+1%2F%283%2A4%29+ ...+1%2Fn%28n%2B1%29+=+n%2F%28n%2B1%29

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1%2F%281%2A2%29+%2B+1%2F%282%2A3%29+%2B+1%2F%283%2A4%29%2B%22...%22%2B1%2F%28n%28n%2B1%29%29%22%22=%22%22n%2F%28n%2B1%29

We substitute n=1 

1%2F%281%2A2%29%22%22=%22%221%2F%281%2B1%29

That's true.

Assume true for n=k

(1)  1%2F%281%2A2%29+%2B+1%2F%282%2A3%29+%2B+1%2F%283%2A4%29%2B%22...%22%2B1%2F%28k%28k%2B1%29%29%22%22=%22%22k%2F%28k%2B1%29

We want to show that if (1) is true, then the following is true:

     %22%22=%22%22%28k%2B1%29%2F%28%28k%2B1%29%2B1%29

which simplifies to this:

(2)  %22%22=%22%22%28k%2B1%29%2F%28k%2B2%29

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So we start with what we assume to be true, which is (1):

1%2F%281%2A2%29+%2B+1%2F%282%2A3%29+%2B+1%2F%283%2A4%29%2B%22...%22%2B1%2F%28k%28k%2B1%29%29%22%22=%22%22k%2F%28k%2B1%29

and add the next term red%281%2F%28%28k%2B1%29%28k%2B2%29%29%29 to both sides:

%22%22=%22%22k%2F%28k%2B1%29%2Bred%281%2F%28%28k%2B1%29%28k%2B2%29%29%29

Now if we can show that the right side

k%2F%28k%2B1%29%2B1%2F%28%28k%2B1%29%28k%2B2%29%29

is equal to the right side of (2) then we will have proved the
proposition by induction:

k%2F%28k%2B1%29%2B1%2F%28%28k%2B1%29%28k%2B2%29%29 =

We get an LCD of (k+1)(k+2)

%28k%28k%2B2%29%29%2F%28%28k%2B1%29%28k%2B2%29%29%2B1%2F%28%28k%2B1%29%28k%2B2%29%29 =

%28k%28k%2B2%29%2B1%29%2F%28%28k%2B1%29%28k%2B2%29%29 =

%28k%5E2%2B2k%2B1%29%2F%28%28k%2B1%29%28k%2B2%29%29 =

%28%28k%2B1%29%28k%2B1%29%29%2F%28%28k%2B1%29%28k%2B2%29%29 =

%28%28cross%28k%2B1%29%29%28k%2B1%29%29%2F%28%28cross%28k%2B1%29%29%28k%2B2%29%29 =

%28k%2B1%29%2F%28k%2B2%29

That's the same as the right side of (2) above.  So we are done.

Edwin