+26387 Hello Guest,
to generate your formula see the link: http://mathworld.wolfram.com/Dice.html
\(\text{The probability of obtaining p points (a roll of p) on n s-sided dice can be computed as follows. $\\$ The number of ways in which p can be obtained is the coefficient of $x^p$ in $\\$ $f(x)=\left(x+x^2+...+x^s \right)^n, \qquad (1)$ } \)
\(\text{since each possible arrangement contributes one term. $f(x)$ can be written as a multinomial series $\\$ $f(x) = x^n \left( \sum \limits_{i=0}^{s-1} x^i \right)^n \qquad (2) \\ = x^n\left(\dfrac{1-x^s}{1-x}\right)^n, \qquad (3)$}\)
\(\text{so the desired number c is the coefficient of $x^p$ in $\\$ $x^n(1-x^s)^n(1-x)^{-n}. \qquad (4)$}\)
\(\text{Expanding,$\\$ $x^n \sum \limits_{k=0}^n (-1)^k \dbinom{n}{k} x^{sk} \sum \limits_{l=0}^{\infty} \dbinom{n+l-1}{ l}x^l, \qquad (5)$ }\)
\(\text{so in order to get the coefficient of $x^p$, include all terms with $\\$ $p=n+sk+l. \qquad (6)$ }\)
\(\text{c is therefore $\\$ $c= \sum \limits_{k=0}^n (-1)^k \dbinom{n}{k} \dbinom{p-sk-1}{p-sk-n}. \qquad (7)$ } \)
\(\text{But $p-sk-n>0$ only when $k<(p-n)/s$, so the other terms do not contribute. Furthermore, $\\$ $\dbinom{p-sk-1}{ p-sk-n}= \dbinom{p-sk-1}{ n-1}, \qquad (8)$}\)
\(\text{so $\\$ $c= \sum \limits_{k=0}^{\Big\lfloor \dfrac{p-n}{s} \Big\rfloor} (-1)^k \dbinom{n}{ k} \dbinom{p-sk-1}{ n-1}, \qquad (9)$}\)
\(\text{where $\lfloor x \rfloor$ is the floor function, and $\\$ $\mathbf{P(p,n,s)= \dfrac{1}{s^n} \sum \limits_{k=0}^{\Big\lfloor \dfrac{p-n}{s} \Big\rfloor} (-1)^k \dbinom{n}{ k} \dbinom{p-sk-1}{ n-1}} \qquad (10)$}\)
+26387 The Fibonacci numbers 1, 1, 2, 3, 5, 8, . . . form a sequence where each term, after the first two, is the sum of the two previous terms.
How many of the first 1000 terms are even?
Even Fibonacci numbers: \(\large f_{3n} ,\ n\in N\)
source: http://oeis.org/search?q=even+fibonacci&sort=&language=german&go=Suche
\(\begin{array}{|rcll|} \hline f_{\color{red}3}=2,\ f_{\color{red}6}=8,\ f_{\color{red}9} = 34,\ f_{\color{red}12}=144,\ \ldots,\ f_{\color{red}3n},\ \ldots \text{ Fibonacci numbers are even} \\ \hline \\ \begin{array}{rcll} \text{arithmetic series:} \\ 3+(n-1)\cdot 3 &=& 1000 \\ (n-1)\cdot 3 &=& 997 \\ n-1 &=& \dfrac{997}{3} \\ n &=& 1+\dfrac{997}{3} \\ n &=& 333.\overline{3} \\ \mathbf{n} &=& \mathbf{333} \\ \end{array} \\ \hline \end{array}\)
