Zadanie:
Oblicz \(\displaystyle{ \sum_{k=1}^{n-1} {n \choose k} \left( -2\right)^k}\)
Rozwiązanie:
Wzór dwumianowy Newtona
\(\displaystyle{ \left( a+b\right)^n= \sum_{k=0}^{n} {n \choose k} a^k b^{n-k}}\)
takim razie jeżeli \(\displaystyle{ a=-2 \wedge b =1}\), to
\(\displaystyle{ \left( -1\right)^n= \left( -2+1\right)^n= \sum_{k=0}^{n} {n \choose k} \left( -2\right) ^k 1^{n-k}= \sum_{k=0}^{n} {n \choose k} \left( -2\right) ^k}\)
\(\displaystyle{ ={n \choose 0} \left( -2\right) ^0+\sum_{k=1}^{n-1} {n \choose k} \left( -2\right) ^k+
{n \choose n} \left( -2\right) ^n}\)
stąd
\(\displaystyle{ \left( -1\right) ^n=1+\sum_{k=1}^{n-1} {n \choose k} \left( -2\right) ^k+\left( -2\right)^n}\)
Ostatecznie:
\(\displaystyle{ \sum_{k=1}^{n-1} {n \choose k} \left( -2\right) ^k=\left( -1\right)^n+\left( -2\right) ^{n+1} -1}\)
Poproszę o komentarz.
Rozwinięcie dwumianowe
Rozwinięcie dwumianowe
\(\displaystyle{ 1+ \sum_{k=1}^{n-1} {n \choose k} (-2)^k +(-2)^n = \sum_{k=0}^{n} {n \choose k} (-2)^k = \sum_{k=0}^{n} {n \choose k} (-2)^k 1^{n-k} =(1-2)^n =(-1)^n}\)
więc
\(\displaystyle{ \sum_{k=1}^{n-1} {n \choose k} (-2)^k =-1 +(-1)^n -(-2)^n}\)
więc
\(\displaystyle{ \sum_{k=1}^{n-1} {n \choose k} (-2)^k =-1 +(-1)^n -(-2)^n}\)