Question 1006222
<pre>
First we show that the expression 

{{{expr(n/2)(n+1)}}}

gives the sum of the first 2 natural numbers:

1+2 = 3

and the expression with n=2 substituted gives:

{{{expr(2/2)(2+1)=(1)(3) = 3}}}

So the formula holds for n=k=22

Now we know that there is at least one natural number k=2 for which
the equation holds for n=k.

Next we show that under the assumption that we just showed, that 
there exists one natural number n=k for which the equation holds true, 
then the equation will also hold for n=k+1

Under the assumption that the expression gives the sum of the first
n=k natural numbers for some n=k, then

{{{expr(k/2)(k+1)}}}

We add the next natural number (k+1) to the expression:

{{{expr(k/2)(k+1)+(k+1)}}}

We factor out (k+1)

{{{(k+1)(k/2+1)=(k+1)(k/2+2/2)=(k+1)((k+2)/2)=(k+1)(1/2)(k+2)=expr((k+1)/2)(k+2)}}}

And this equals to the expression 

{{{expr(n/2)(n+1)}}} with k+1 substituted for n, since

{{{expr((k+1)/2)((k+1)^""+1)=expr((k+1)/2)(k+2)}}}

Now since we have shown that it is true when n=k=2, it is therefore
true when n=k+1=3.

Now since we have shown that it is true when n=k=3, it is therefore
true when n=k+1=4.

Etc., etc., 

Therefore there can be no first value of k for which the expression 
does not hold.  For if there were such first value, the expression
would hold for n=k-1 and therefore it would hold for n=k, which would 
be a contradiction to the assumption that there could be a natural 
number k for which the expression did not hold.

Edwin</pre>