SOLUTION: The number 3 can be expressed as a sum of one or more positive integers in four ways, namely, as 3, 1+2, 2+1, and 1+1+1. Show that any positive integer n can be so expressed in 2<s

Think of those 4 ways as ways to group a sum of 3 1's:

3     = (1 + 1 + 1)
1+2   = (1)+(1 + 1)
2+1   = (1 + 1)+(1)
1+1+1 = (1)+(1)+(1)

We start with the longest possible petition

3     =   (1 + 1 + 1)

And we can either choose to replace either of the 2 
" + " signs by ")+(" or leave it as " + ".  So starting
with (1 + 1 + 1) we have 2 choices to make with each +
sign,

Either
1. We may choose to leave it as " + "
Or
2. We may choose to change it to ")+("

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For the general case, suppose we have the integer n.  Then write 
it as the sum of n 1's all in one set of parentheses, like this:

n = (1 + 1 + 1 + ··· + 1 + 1 + 1)

There are n-1 plus signs (because there is a + sign
after each 1 except there isn't a + after the last one).

For each of those n-1 plus signs,
we have two possible choices to make:

1. We may choose to leave it as " + "
OR
2. We may choose to change it to ")+("

So there are 2 choices to make for the 1st + sign.
For each of those 2 choices we can make for the 1st + sign, 
there are 2 choices we can make for the 2nd  + sign.
That's 22 choices for the first 2 + signs.
For each of those 22 choices we can make for the 1st 2 + signs, 
there are 2 choices we can make for the 3rd  + sign.
That's 23 choices for the first 3 + signs.
...

this continues, and since we have (n-1) plus signs,

the answer is 2n-1 ways of leaving the n-1
+ signs as they are or changing them to ")+(".

Edwin