Question

find the next term

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{Find the next term}\hspace{.33em}\\~\\
& 1,\ \ 3,\ \ 7,\ \ 25,\ \ 103,\ \ ?\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

1,3,7,25,103
1,4,18,78
3,14,60
11,46
doesn't look like we have enough terms to figure out if it is a polynomial
let me look for a different pattern

Answer 3

and i made a mistype but it still doesn't help

Answer 4

answer given is \(521\) but i didnt understand

Answer 5

where did you get this question

Answer 6

wait let me post the reasoning given

Answer 7

i found this on a youtube video

Answer 8

and i guess your question is where the 2,1,4,3,6 are coming from?

Answer 9

^ yep

Answer 10

the other numbers are definitely obvious

Answer 11

at first it looked like term-previous term
but then that changed when we got to the third equation

Answer 12

i mean the fourth

Answer 13

i suppose this is a faulty sequence unless i m missing something

Answer 14

i think I got something

Answer 15

\[a_n \cdot n+b_n=a_{n+1}\]
so this is the equation we have
where we still need to figure out b(n)

Answer 16

b(n) goes like this
2,1,4,3,6,...
find first differences -1,3,-1,3,...
notice the pattern?

Answer 17

yes :D

Answer 18

so let me split this sequence up
2,4,6,...
1,3,5,7,....

Answer 19

like you have even every other entry
and odd every other entry

Answer 20

\(\Huge \bf ☺\ \ \ \) i thought this was impossible to solve

Answer 21

having a dumb moment right now to set up something general
but anyways
the next number in 2,1,4,3, is 6
which is how we get 521

Answer 22

\(\Huge 😺\) m happy

Answer 23

\[2,1,4,3,6,.... \\ b_1=2,b_3=4,b_5=6, \cdots , b_{2k+1}=2(k+1) \\ k \in \mathbb{Z}^{+,0} \\ \\ b_2=1,b_4=3, \cdots , b_{2k}=2k-1\]

Answer 24

thnks

Answer 25

\[b_n=a_{n+1}-a_n \cdot n \\ \text{ if } n \text{ is odd } \\ \text{ then we have } \\ b_{2k+1}=a_{(2k+1)+1}-a_{2k+1} \cdot (2k+1) \\ \text{ then w ehave } \\ 2(k+1)=a_{2(k+1)}-a_{2k+1} \cdot (2k+1) \\ \text{ and if } n \text{ is even } \\ b_{2k}=a_{2k+1}-a_{2k} \cdot (2k) \\ 2k-1=a_{2k+1}-a_{2k} \cdot (2k)\]

ok now I'm I think and I will stop replying

Answer 26

i just wanted to type something general

Answer 27

u mean the general term

Answer 28

but this question is from logical reasoning

Answer 29

the general recurrence relation

Answer 30

it looks like it
but it is not easy to see
i think it would be harder if you didn't know that one term was 521
like the pattern would be harder to see maybe

Answer 31

yes without knowing 521 i think it is impossible

Answer 32

for people like me

Answer 33

even after i know 521

Answer 34

i probably wouldn't even guess the recurrence relation without the equations you gave... :p

Answer 35

so don't feel bad

Answer 36

ok

Answer 37

I want to state it this way
I think this is more readable
\[a_n-n=a_{n-1}(n-1) \text{ if } n \in \{2,4,6,... \} \\ a_n-(n-2)=a_{n-1}(n-1) \text{ if } n \in \{3,5,7,... \}\]