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# Find out the wrong number in the given sequence of

## Tags: Mathematik

23:28 Uhr, 23.11.2022

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,

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

23:42 Uhr, 23.11.2022

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

23:56 Uhr, 23.11.2022

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

09:11 Uhr, 24.11.2022

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 $n\phantom{\rule{0.12em}{0ex}}=1,2,\dots ,7$

21:17 Uhr, 24.11.2022

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 $3\left(1.5\right)$ 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

21:48 Uhr, 24.11.2022

> when i take the half of 3(1.5) does not work

It's sad that you don't read the postings. I repeat: $f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)=\frac{3{n\phantom{\rule{0.12em}{0ex}}}^{2}+n\phantom{\rule{0.12em}{0ex}}+12}{2}$

00:39 Uhr, 25.11.2022

$i\phantom{\rule{0.12em}{0ex}}$ 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.

01:46 Uhr, 25.11.2022

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.

10:20 Uhr, 25.11.2022

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":

$f\phantom{\rule{0.12em}{0ex}}\left(n\phantom{\rule{0.12em}{0ex}}\right)=8+5\cdot \left(n\phantom{\rule{0.12em}{0ex}}-1\right)+\frac{3}{2}\cdot \left(n\phantom{\rule{0.12em}{0ex}}-1\right)\left(n\phantom{\rule{0.12em}{0ex}}-2\right)=\frac{3{n\phantom{\rule{0.12em}{0ex}}}^{2}+n\phantom{\rule{0.12em}{0ex}}+12}{2}$

See en.wikipedia.org/wiki/Newton_polynomial#Application for details.

15:48 Uhr, 25.11.2022

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

15:48 Uhr, 25.11.2022

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