How do you prove #1/(1*2)+1/(2*3)+...+1/(n(n+1)) = 1-1/(n+1)# by induction?
3 Answers
See explanation...
Explanation:
Let
#1/(1*2)+1/(2*3)+...+1/(n(n+1)) = 1-1/(n+1)#
Base case
Note that:
#1/(1*2) = 1/2 = 1-1/2 = 1-1/(1+1)#
So
Induction step
Suppose
Then:
#1/(1*2)+1/(2*3)+...+1/(n(n+1))+1/((n+1)(n+2)#
#=1-1/(n+1)+1/((n+1)(n+2))#
#=1-((n+2)/((n+1)(n+2))-1/((n+1)(n+2)))#
#=1-((n+2)-1)/((n+1)(n+2))#
#=1-(n+1)/((n+1)(n+2))#
#=1-1/(n+2)#
That is
Conclusion
We have shown
So by induction
Show that it is true for
Explanation:
In mathematical induction, there are two steps:
1. Show that it is true for the first term
2. Show that if it is true for a term
Here is our current sequence:
Huh, this feels a bit wrong, there's a similar answered question, but it's not quite the same. Anyway, we should be able to rewrite this sum in terms of sigma:
The first step is to show that it is true for the first term. Here we go,
Nice! Turns out this was a different sum than what I thought. For the second step, assume that the sum is true for a constant, say,
And now, we need to prove that it is true for
Let's start a little proof by contradiction: assume it isn't, then show that it must be.
The thing is, though, is that
At this point we can start ignoring
We can subtract both sides by
Interesting. Let's somehow make all of these fractions share the same denominator:
Combining those two fractions on the left hand side:
Subtracting the fraction on the right hand side, by itself:
Let's simplify:
We have a contradiction! Therefore, it is also true that if the sequence is true for a term
Proof via method of differences
Explanation:
I wanted to provide another approach to showing this is true:
We know
Using partial fractions:
Letting
Letting
.
.
.
Hence using what is left: