**show that C(n+1,k)=C(n,k-1)+C(n,k)**

The **Answer to the Question**

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**Here's the Solution to this Question**

$C_{n + 1}^k = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}}$

$C_n^{k - 1} + C_n^k = \frac{{n!}}{{(k - 1)!(n - k + 1)!}} + \frac{{n!}}{{k!(n - k)!}} = \frac{{n!k}}{{k!(n - k + 1)!}} + \frac{{n!(n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!k + n!(n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!(k + n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!(n + 1)}}{{k!(n - k + 1)!}} = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}}$

$\frac{{(n + 1)!}}{{k!(n + 1 - k)!}} = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}} \Rightarrow C_{n + 1}^k = C_n^{k - 1} + C_n^k$