If $a_n = \frac{-2}{4n^2 - 16n + 15}$,then $a_1 + a_2 + \dots + a_{25}$ is equal to:

The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$, then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

If $t_{n} = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \dots$,then find the value of $\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \dots + \frac{1}{t_{2003}}$.

Statement-$1$: The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)+\dots+(361+380+400)$ is $8000$.
Statement-$2$: $\sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3$,for any natural number $n$.

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to