Not Pascal's Identity

Note first that for $0 \le n \le 14$ we have that

$\dfrac{C_{n+1}}{C_{n}} = \dfrac{\frac{15!}{(14 - n)! * (n + 1)!}}{\frac{15!}{(15 - n)! * n!}} =\dfrac{15 - n}{n + 1}$ .

So the desired sum is

$\displaystyle\sum_{n=0}^{14} (n + 1)\dfrac{(15 - n)}{n + 1} = 15*\displaystyle\sum_{n=0}^{14} (1) - \sum_{n=0}^{14} n = 15*15 - \dfrac{14*15}{2} = \boxed{120}$ .

Note that f(k) = (k+1) * 15C(k+1)/15Ck = (k+1)(k!)(15-k)! / (k+1)!(14-k)! = 15-k. So the required sum is 15+14+...+1 = 15(15+1)/2 = 120.

C(r)/C(r-1) = (n+1)/r - 1 And T(r) = r * C(r)/C(r-1) = (n+1) - r Thus Sum = n(n+1) - n(n+1)/2 = n(n+1)/2 Putting n = 15, you get 120

Since $×$ is between 1 and 15, inclusive, l simplified the fraction ${ 15 \choose x} / _{15 \choose (x-1) }$ which came out to $(16-x)/x$ . Thus, the problem can be rewritten as $\sum_{i =1}^{15} x((16-x)/x) = \sum_{i=1}^{15} (16-x) = (15 \times 16)/2 = 15 \times 8 = 120$

15*14=210=>120

$\frac{C_{x}}{C_{x-1}}=\frac{\frac{15!}{x!(15-x)!}}{\frac{15!}{(x-1)!(15-x+1)!}}=\frac{16-x}{x}$

$\frac {C_1}{C_0} + 2\ \frac {C_2}{C_1} + 3\ \frac {C_3}{C_2} + \ldots + 15\ \frac {C_{15}}{C_{14}} \\ =\sum_{x=1}^{15}x\frac{C_{x}}{C_{x-1}}=\sum_{x=1}^{15}x\frac{16-x}{x}=\sum_{x=1}^{15}{16-x}=\sum_{x=1}^{15}{16}-\sum_{x=1}^{15}{x}=15\times16-\frac{15\times16}{2}=\frac{15\times16}{2}=\boxed{120}$