My Version of Pascal's Triangle

Let $\left\{ \begin{matrix} n \\ p \end{matrix} \right\}$ the element in the $n$ -th row and $p$ -th column. So $\left\{ \begin{matrix} n \\ 0 \end{matrix} \right\} = \left\{ \begin{matrix} n \\ p \end{matrix} \right\} = n$ and the "Pascal Rule" holds: $\left\{ \begin{matrix} n \\ p \end{matrix} \right\} + \left\{ \begin{matrix} n \\ p+1 \end{matrix} \right\} = \left\{ \begin{matrix} n+1 \\ p+1 \end{matrix} \right\}$ Then, we have: $f(n) = \sum_{p=0}^{n} \left\{ \begin{matrix} n \\ p \end{matrix} \right\} \Rightarrow$ $2f(n) = \left\{ \begin{matrix} n \\ 0 \end{matrix} \right\} + \sum_{p=0}^{n-1} \left[ \left\{ \begin{matrix} n \\ p \end{matrix} \right\} + \left\{ \begin{matrix} n \\ p + 1 \end{matrix} \right\} \right] + \left\{ \begin{matrix} n \\ n \end{matrix} \right\} \Rightarrow$ $2f(n) + 2 = \sum_{p=0}^{n+1} \left\{ \begin{matrix} n +1\\ p+1 \end{matrix} \right\} \Rightarrow$ $f(n+1) = 2f(n) + 2$ Notice that $f(0) = 0$ . Then: $\sum_{k=1}^{n} \dfrac{f(k)}{2^{k-1}} = \sum_{k=1}^{n} \dfrac{f(k-1)}{2^{k-2}} +\dfrac{1}{2^{k-2}} \Rightarrow$ $f(n) = 2^{n+1} -2$

So, $\dfrac{f(100)}{f(50)} = \dfrac{2^{101 }-2}{2^51 - 2} = 2^50+1$ . Notice that $2^{10} = 1024 \equiv -1 \mod 25 \Rightarrow 2^{50} + 1 \equiv 0 \mod 25$ So, the answer is $0$

Nice problem but overrated!

$f(0) = 0 \ , \ f(1) = 2 \ , \ f(2) = 6 \ , \ f(3) = 14 \Rightarrow\ f(n) = 2(2^n-1)$ Now we assume this is true and use induction. Well it is clear to see that each element in a row except for the ends is counted twice to obtain the next row. This means we have to over-count, subtract the elements at the end and then add the ends of the next row as follows: $f(n+1) = 2f(n) -2n +2(n+1) = 4(2^n-1) +2$ $= 2^{n+2} - 2 = 2(2^{n+1} -1) \ as \ required$ SO: $\frac{f(100)}{f(50)} = \frac{2(2^{100} -1)}{2(2^{50}-1)} = \frac{(2^{50} +1)(2^{50}-1)}{2^{50} -1} = 2^{50} +1$ $2^{10} \equiv-1\mod(25)$ $\therefore\ 2^{50} + 1\equiv-1+1\equiv0\mod(25)$

By observation and induction it can be shown that (see this special Pascal Triangle above):

$f(n) = 2^{n+1}-2$

Therefore,

$\dfrac {f(100)}{f(50)} = \dfrac {2^{101}-2}{2^{51}-2} = \dfrac {2^{100}-1}{2^{50}-1} = \dfrac {(2^{50}+1)(2^{50}-1)}{2^{50}-1} = 2^{50}+1$

Now it is to find:

$(2^{50}+1) \pmod {25} \equiv [(2^{10})^5 + 1] \pmod{25}$

$\equiv [(1025-1)^5 + 1] \pmod {25} \equiv (-1+1) \pmod {25} \equiv \boxed{0} \pmod {25}$