I proved each of the following via mathematical induction.

1 1·2

+ 
1 2·3

+ ¼+ 
1 n· (n+1)

= 1  
1 n+1

for n ³ 1 . 

The format for these proofs is standard. Here is a proof of the third:
The base case is n = 0.
We check whether 6  0^{3} 0, i.e.,
whether 6  0 and see that the answer is yes.
Now assume that for some k, 6  k^{3} k.
We need to prove
that 6  (k+1)^{3} (k+1).
But (k+1)^{3} (k+1) = (k^{3}  k)+3(k^{2}+k). Since 6  k^{3} k, we need only show that
6  3(k^{2}+k). This follows provided 2  k^{2}+k. But
k^{2}+k = k(k+1) and either k of k+1 must be even so we are done.
As an illustration of what can happen if one is careless in applying
the assumption, consider the following ``proof'' that all positive
numbers are equal.
We will prove that if max(a,b) = n then a = b. If n = 1 both a
and b must be 1, hence a = b. Assume that for some k, we have
max(a,b) = k Þ a = b. If max(a,b) = k+1 then
max(a1,b1) = k from which it follows that a1 = b1 so that
a = b.
The problem is (as it turns out) that there is no guarantee
that a1 and b1 are positive integers.
Next time I will consider induction where rather than
assuming that the property holds for k (and trying to prove it holds
for k+1) we will assume that the property holds for all numbers less
than or equal to k but otherwise proceed as above.
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On 30 Oct 2000, 19:43.