If $$\frac{1}{1 \times 3}+\frac{1}{2 \times 4}+\frac{1}{3 \times 5}+\cdots+\frac{1}{n(n+2)}=\frac{3553}{4830}$$, then $$n$$ is

Notice that

$$\dfrac{1}{k(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right).$$

Using this, the given sum becomes

$$\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\cdots+\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right).$$

Most of the terms cancel out, leaving only

$$\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right).$$

Therefore,

$$\dfrac{1}{1 \times 3}+\dfrac{1}{2 \times 4}+\cdots+\dfrac{1}{n(n+2)}=\dfrac{3}{4}-\dfrac{2n+3}{2(n+1)(n+2)}.$$

We are given that this sum is $$\dfrac{3553}{4830}.$$
We can substitute the values and check.

For $$n=68$$.

$$\dfrac{3}{4}-\dfrac{2(68)+3}{2(69)(70)}=\dfrac{3}{4}-\dfrac{139}{9660}=\dfrac{3553}{4830}.$$

This matches the given value.