Welcher Term besteht zwischen 1,12,33und 62
\\ \boxed{p(n) = a*n^3+b*n^2+c*n+d} \\ p(1)= 1 = a*1^3+b*1^2+c*1+d \\ p(2)= 12 = a*2^3+b*2^2+c*2+d \\ p(3)= 33 = a*3^3+b*3^2+c*3+d \\ p(4)= 62 = a*4^3+b*4^2+c*4+d \\ \hline (1)\ a+b+c+d=1 \\ (2)\ 8a+4b+2c+d=12\\ (3)\ 27a+9b+3c+d=33\\ (4)\ 64a+16b+4c+d=62\\ \hline I: (2)-(1): \ 7a+3b+c=11 \\ II:(4)-(3): \ 37a+7b+c+d=29\\ III: (3)-(2): \ 19a+5b+c=21\\ \hline
III-I: \ 12a+2b=10\\ II-III: \ 18a+2b =8 \\ 18a-12a=8-10 \\ 6a=-2\\ a=-\frac{1}{3}\\ -\frac{12}{3}+2b=10\\ 2b=10+4\\ b=7\\ \hline -\dfrac{7}{3}+21+c=11\\ c=-\dfrac{23}{3}\\ \hline -\dfrac{1}{3}+7-\dfrac{23}{3}+d=1\\ -\dfrac{24}{3}+7+d=1 \\ d=2 \\ \hline
p(n)=−13∗n3+7∗n2−233∗n+2p(1)=−8+7+2=1 okay!p(2)=30−18=12 okay!p(3)=65−32=33 okay!p(4)=114−52=62 okay!
Welcher Term besteht zwischen 1,12,33und 62
\\ \boxed{p(n) = a*n^3+b*n^2+c*n+d} \\ p(1)= 1 = a*1^3+b*1^2+c*1+d \\ p(2)= 12 = a*2^3+b*2^2+c*2+d \\ p(3)= 33 = a*3^3+b*3^2+c*3+d \\ p(4)= 62 = a*4^3+b*4^2+c*4+d \\ \hline (1)\ a+b+c+d=1 \\ (2)\ 8a+4b+2c+d=12\\ (3)\ 27a+9b+3c+d=33\\ (4)\ 64a+16b+4c+d=62\\ \hline I: (2)-(1): \ 7a+3b+c=11 \\ II:(4)-(3): \ 37a+7b+c+d=29\\ III: (3)-(2): \ 19a+5b+c=21\\ \hline
III-I: \ 12a+2b=10\\ II-III: \ 18a+2b =8 \\ 18a-12a=8-10 \\ 6a=-2\\ a=-\frac{1}{3}\\ -\frac{12}{3}+2b=10\\ 2b=10+4\\ b=7\\ \hline -\dfrac{7}{3}+21+c=11\\ c=-\dfrac{23}{3}\\ \hline -\dfrac{1}{3}+7-\dfrac{23}{3}+d=1\\ -\dfrac{24}{3}+7+d=1 \\ d=2 \\ \hline
p(n)=−13∗n3+7∗n2−233∗n+2p(1)=−8+7+2=1 okay!p(2)=30−18=12 okay!p(3)=65−32=33 okay!p(4)=114−52=62 okay!