Question

write the recursive formular in explicit form \[a_n=a_{n-1}r+d\]

Answer 1

for GP \[a_n=a_{n-1}r\implies a_n=a_0r^n\]
for AP
\[a_n=a_{n-1}+d\implies a_n=a_0+(n-1)d\]
i want to combine this two

Answer 2

example \[a_n=4a_{n-1}+1\]

Answer 3

woah thats complicated

Answer 4

i will try to compute it manually
\[a_0=a\] is given
\[a,4a+1,4(4a+1)+1,4(4(4a+1)+1)+1,4(4(4(4a+1)+1))+1,4(4(4(4(4a+1)+1))+1)+1\]
this is what it gives if we consider the bracket openings we see that
\[a,4a+1,4^2a+4,4^3a+4^2+4+1,4^4a+4^3+4^2+4+1,....4^ka+4^{k-1}+...+4^2+4+1\]
so we can say generally
\[a_n=4^na+4^{n-1}+...+4^2+4+1=4^na+\frac{4^n-1}{3}\]

Answer 5

and more generally
\[\huge a_n=ar^n+d\frac{r^n-1}{r-1}\]

Answer 6

i have no idea whats going on :( i wish i could help

Answer 7

lol thank you @plohrr

Answer 8

:D

Answer 9

if we repeateadly use the same formular we have
\[a_n=a_{n-1}r+d=(a_{n-2}r+d)r+d=a_{n-2}r^2+dr+d=a_{n-3}r^3+dr^2+dr+d\\a_n=\color{blue}{a_0r^n+d\frac{1-r^n}{1-r}}\]