To find the sum of the series 1/2 + 1/6 + 1/12 + ... + 1/x(x+1), we need to first find a pattern in the terms of the series.

The general term of the series can be written as 1/(n(n+1)), where n takes on values from 1 to x.

Now, we can rewrite the series as:

1/2 + 1/6 + 1/12 + ... + 1/x(x+1) = Σ(1/(n(n+1)), n=1 to x)

Next, we need to find a common denominator for all the terms in the series. The common denominator for the series is x(x+1).

Now, we can rewrite each term with the common denominator:

1/2 = x(x+1)/(2x(x+1))
1/6 = x(x+1)/(6x(x+1))
1/12 = x(x+1)/(12x(x+1))
...
1/x(x+1) = x(x+1)/(x(x+1))

Now, we can rewrite the series with the common denominator:

Σ(1/(n(n+1)), n=1 to x) = Σ(x(x+1)/(nx(x+1)), n=1 to x)

Next, we can simplify the series by canceling out the common terms:

Σ(x(x+1)/(nx(x+1)), n=1 to x) = Σ(1/n, n=1 to x)

Therefore, the sum of the series 1/2 + 1/6 + 1/12 + ... + 1/x(x+1) is equal to the sum of the harmonic series from 1 to x, which is equal to ln(x+1).