Symbol Newtona rozwiąż równania.

[bcm]

Potrzebuje pomocy w rozwiązaniu poniższych równań.

1. \(\displaystyle{ {n+3 \choose n} =20}\)

2. \(\displaystyle{ {n \choose n-2} + {n+4 \choose 2} = 24}\)

3. \(\displaystyle{ k \cdot {n \choose k} = n {n-1 \choose k-1}}\)

4. \(\displaystyle{ {2n \choose 2} = 2 \cdot {n \choose 2} + n^{2}}\)

[kerajs]

1.
\(\displaystyle{ {n+3 \choose n} =20}\)
\(\displaystyle{ \frac{(n+1)(n+2)(n+3)}{3!}=20 \\ (n+1)(n+2)(n+3)=4 \cdot 5 \cdot 6\\n=3}\)

2.
\(\displaystyle{ {n \choose n-2} + {n+4 \choose 2} = 24}\)
\(\displaystyle{ \frac{(n-1)(n)}{2!} + \frac{(n+3)(n+4)}{2!}=24 \\ (n-1)(n)+(n+3)(n+4)=2 \cdot 3 +6\cdot7\\n=3}\)

3.
\(\displaystyle{ k \cdot {n \choose k} = n {n-1 \choose k-1}}\)
\(\displaystyle{ L=k \cdot {n \choose k} =k \frac{n!}{k!(n-k)!}= \frac{(n-1)!n}{(k-1)!(n-k)!}= \\=n \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}= n {n-1 \choose k-1}=P}\)

4.
\(\displaystyle{ {2n \choose 2} = 2 \cdot {n \choose 2} + n^{2}}\)
\(\displaystyle{ L={2n \choose 2} = \frac{(2n-1)2n}{2}=n(2n-1)=n^2-n+n^2 =\\=2 \frac{(n-1)n}{2}+n^2=
2 \cdot {n \choose 2} + n^{2}=P}\)

Washington na zbożny cel.