$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots$ to $24$ terms $=$

The sum to $50$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is:

If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$,then the maximum value of $a$ is.

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

Find the sum to $n$ terms of the series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots$