Zbiór zadań maturalnych/Matematyka/Zadania/R1

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Zadanie R1 (0-2)
Wykaż, że ${\binom {n}{k}}+{\binom {n}{k+1}}={\binom {n+1}{k+1}}$

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Zasady oceniania
Przykładowe rozwiązania	Niech ${\binom {n}{k}}+{\binom {n}{k+1}}=a$ oraz ${\binom {n+1}{k+1}}=b$ wtedy ${\begin{aligned}a=&\ {\binom {n}{k}}+{\binom {n}{k+1}}&\\[10pt]a=&\ {\frac {n!}{(n-k)!\cdot k!}}+{\frac {n!}{[n-(k+1)]!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!}{(n-k)!\cdot k!}}+{\frac {n!}{(n-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot [(n-k-1)!\cdot (k+1)!]}{[(n-k)!\cdot k!]\cdot [(n-k-1)!\cdot (k+1)!]}}+{\frac {[(n-k)!\cdot k!]\cdot n!}{[(n-k)!\cdot k!]\cdot [(n-k-1)!\cdot (k+1)!]}}&\\[10pt]a=&\ {\frac {n!\cdot [(n-k-1)!\cdot (k+1)!]+[(n-k)!\cdot k!]\cdot n!}{[(n-k)!\cdot k!]\cdot [(n-k-1)!\cdot (k+1)!]}}&\\[10pt]a=&\ {\frac {n!\cdot [(n-k-1)!\cdot (k+1)!]+[(n-k)!\cdot k!]\cdot n!}{(n-k)!\cdot k!\cdot (n-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot [(n-k-1)!\cdot (k+1)!]+[(n-k)\cdot (n-k-1)!\cdot k!]\cdot n!}{(n-k)!\cdot k!\cdot (n-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot \{1\cdot [(n-k-1)!\cdot (k+1)!]+[(n-k)\cdot (n-k-1)!\cdot k!]\cdot 1\}}{(n-k)!\cdot k!\cdot (n-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot (n-k-1)!\cdot \{1\cdot [1\cdot (k+1)!]+[(n-k)\cdot 1\cdot k!]\cdot 1\}}{(n-k)!\cdot k!\cdot (n-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot {\cancel {(n-k-1)!}}\cdot \{{\cancel {1}}\cdot [{\cancel {1}}\cdot (k+1)!]+[(n-k)\cdot {\cancel {1}}\cdot k!]\cdot {\cancel {1}}\}}{(n-k)!\cdot k!\cdot {\cancel {(n-k-1)!}}\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot \{[(k+1)\cdot k!]+[(n-k)\cdot k!]\}}{(n-k)!\cdot k!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot k!\{[(k+1)\cdot {\cancel {1}}]+[(n-k)\cdot {\cancel {1}}]\}}{(n-k)!\cdot k!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot {\cancel {k!}}[(k+1)+(n-k)]}{(n-k)!\cdot {\cancel {k!}}\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot [{\cancel {k}}+1+n-{\cancel {k}}]}{(n-k)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {n!\cdot (n+1)}{(n-k)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {(n+1)!}{(n+1-k-1)!\cdot (k+1)!}}&\\[10pt]a=&\ {\frac {(n+1)!}{[(n+1)-(k+1)]!\cdot (k+1)!}}&\\[10pt]a=&\ {\binom {n+1}{k+1}}=b&\ \\[10pt]&&{\text{co kończy dowód}}\\[10pt]\end{aligned}}$
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