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**Deci** For induction what I did is $\displaystyle 4^n + 14 = 6N$

for n = 1

$\displaystyle 4^1 + 14 = 18 = 6 x 3$

for n = k

$\displaystyle 4^k + 14 = 6N$

for n = k+1

$\displaystyle 4^{k+1} + 14 = 6N$

$\displaystyle (4^{k} + 14) + 4^{k+1} + 14 = 6N$

$\displaystyle 2(4^k) +18 = 6N$

and for (2)

since $\displaystyle \left(\begin{array}{cc}n\\r\end{array}\right) = \frac{n!}{r! (n-r)!} $,

so

$\displaystyle \left(\begin{array}{cc}n\\r\end{array}\right) \left(\begin{array}{cc}n\\r+1\end{array}\right) = $

$\displaystyle \frac{n!}{r! (n-r)!} \frac{n!}{(r+1)! (n-(r+1))!} $

$\displaystyle = \frac{n!}{r! (n-r)!} \frac{n!}{(r+1)! (n-r-1)!} $

$\displaystyle = \frac{n! n!}{(r+1)! r! (n-r)! (n-r-1)!} $