I got the following problem and can't solve it.
Factorize the following statement: C(n+2, n) + C(n+3, n+2). So basically they are asking to rewrite the expression as a X*P expression instead of A+B expression.
I think it has to do something with Pascals formula: C(n, k) = C(n-1, k) + C(n-1, k-1).
Regards!
On
You can use these basic identities, $$\binom{n}{k} = \binom{n}{n-k}\ \ \ \ \ \ \binom{n}{1} = n\ \ \ \ \ \ \binom{n}{2} = \frac{n(n-1)}{2}$$ Therefore, $$ \binom{n+2}{n} + \binom{n+3}{n+2} = \binom{n+2}{(n+2)-n} + \binom{n+3}{n+3 -(n+2) } $$ $$ = \binom{n+2}{2} + \binom{n+3}{1}$$ The last one equality ends as suggest @Dr.Sonnhard, but because if you can to use in "somewhere" the identity of Pascal (I don't know why), I'd use in the next way $$ = \binom{n+2}{2} + \binom{n+3}{1} =\binom{n+2}{2} + \left(\binom{n+2}{0}+\binom{n+2}{1}\right) $$ $$ = \binom{n+2}{0} + \binom{n+2}{1}+\binom{n+2}{2} $$ By definition: $$ = (n+2)!\left(\frac{1}{0!(n+2)!} + \frac{1}{1!(n+1)!} + \frac{1}{2!n!}\right) $$ And, it's a expression with form $x*y$
Btw, answer will be: $n+\frac{1}{2} (n+1) (n+2)+3$
by definition we get $\frac{(n+2)!}{n!\cdot 2}+\frac{(n+3)!}{1!\cdot (n+2)!}=n+3+\frac{1}{2}(n+1)(n+2)$ and we obtain $\frac{1}{2}(n^2+5n+8)$
Sonnhard.