

Tag! Find out the wrong number in the given sequence of numbers. $8,13,21,32,47,63,83$ Well, I tried to see if this is a geometric sequence but it is not, because I don't see there is a common ratio among the numbers listed. Also, I tried the arithmetic sequence, and found no common difference here, Any lead will be appreciated, thanks in advance!. Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert): "Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen." 

Replace 47 with 46 and then it's a quadratic sequence. Verified by iterated differences: zero order (original sequence): 8,13,21,32,46,63,83 first order: 5,8,11,14,17,20 second order: 3,3,3,3,3 

Thanks a lot, my friend. that is good!.Richtig gut! 

associated quadratic function: $f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)=\frac{3{n\phantom{\rule{0.12em}{0ex}}}^{2}+n\phantom{\rule{0.12em}{0ex}}+12}{2}$ for $n\phantom{\rule{0.12em}{0ex}}=1,2,\dots ,7$ 

Hello, I have one question. I have been trying to find the nth number for this quadratic sequence but I am not getting anywhere, could you help? I know that the first difference is $5+8+11+14+17+20$ and that the second difference is $+3+3+3+3+3$ $i\phantom{\rule{0.12em}{0ex}}$ know this is a quadratic sequence because the second difference is the same, so it should start off at $n\phantom{\rule{0.12em}{0ex}}}^{2$ I have seen videos where the second difference is even. then the tutor says to have the number and put it in front of $n\phantom{\rule{0.12em}{0ex}},$ let's suppose the second difference is $4,$ then it would be $2{n\phantom{\rule{0.12em}{0ex}}}^{2},$ you see, but in this case the second difference is 3 ad when when $i\phantom{\rule{0.12em}{0ex}}$ take the half of $3(1.5)$ does not work how can $i\phantom{\rule{0.12em}{0ex}}$ go about finding the nth term when the second difference is an odd number? thanks for any help 

> when i take the half of 3(1.5) does not work It's sad that you don't read the postings. I repeat: $f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)=\frac{3{n\phantom{\rule{0.12em}{0ex}}}^{2}+n\phantom{\rule{0.12em}{0ex}}+12}{2}$ 

$i\phantom{\rule{0.12em}{0ex}}$ made a mistake. $i\phantom{\rule{0.12em}{0ex}}$ forgot to change the value of $n\phantom{\rule{0.12em}{0ex}}$. I'll fix it and post my work tomorrow. Sorry. 

when $n\phantom{\rule{0.12em}{0ex}}=1\left(f\phantom{\rule{0.12em}{0ex}}1\right)=3{\left(1\right)}^{2}+1+\frac{12}{2}=8$ when $n\phantom{\rule{0.12em}{0ex}}=2f\phantom{\rule{0.12em}{0ex}}\left(2\right)=3{\left(2\right)}^{2}+2+\frac{12}{2}=13$ and so forth and so on until, $n\phantom{\rule{0.12em}{0ex}}=7f\phantom{\rule{0.12em}{0ex}}\left(7\right)=3{\left(7\right)}^{2}+7+\frac{12}{2}=83$ Thanks, Just one doubt . this is a set formula or you come up with it. I have not seen it in the videos $i\phantom{\rule{0.12em}{0ex}}$ have watched $3{n\phantom{\rule{0.12em}{0ex}}}^{2},$ I know, but why $..$. $+n\phantom{\rule{0.12em}{0ex}}+\frac{12}{2}$ the 2 diivides everytihng just in case it doesn;t show properly I will appreciate your answer. thanks a lot for such a good support. 

A direct way to obtain the formula $f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)$ uses the first elements of each row of the iterated differences, i.e. "8 , 5 , 3": $f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)=8+5\cdot (n\phantom{\rule{0.12em}{0ex}}1)+\frac{3}{2}\cdot (n\phantom{\rule{0.12em}{0ex}}1)(n\phantom{\rule{0.12em}{0ex}}2)=\frac{3{n\phantom{\rule{0.12em}{0ex}}}^{2}+n\phantom{\rule{0.12em}{0ex}}+12}{2}$ See en.wikipedia.org/wiki/Newton_polynomial#Application for details. 

Super!. I get it.Das verstehe ich vollkommen!. thanks, Hal! 

Super!. I get it.Das verstehe ich vollkommen!. thanks, Hal! 