Question 933337
a sub n means the "nth" term of the sequence.

Example: 1,2,3,4,5,....,

the nth term is n. We say "a sub n = n"

Example 2: 1,4,9,16,25,...

We say "a sub n = n^2".



1. 53,477,4293,38637,347733,...

to get next term multiply previous one by {{{9}}}

 a sub n ={{{a[n]=53*9^n}}}(for all terms given)
 if n=0,1,2...

check:
{{{a[0]=53*9^0=53}}}

{{{a[1]=53*9^1=477}}}

{{{a[2]=53*9^2=4293}}}

{{{a[3]=53*9^3=38637}}}

{{{a[4]=53*9^4=347733}}}

or
{{{a[n] = 53 *9^(n-1)}}} if n=1,2... 

check:
{{{a[1] = 53 *9^(1-1)=53*9^0=53*1=53}}}
:
:

I prefer this one because is better to have {{{a[1]}}} as first term



2. 2,5,10,17,26,...
pattern

2----  5----  10----  17----  26,..
--   3----  5----    7--   9...............first diff

----     2----   2----   2................second diff.; all {{{2}}}s, so degree is {{{2}}},  half of it (or {{{1}}})will be coefficient of {{{n^2}}} 

use
 {{{an^2 + bn + c}}} , set up system of three equations to find coefficients {{{a}}},{{{b}}}, and {{{c}}}   

For instance, I know that the first term (that is, the term when n = 1) is {{{2}}}, so I'll plug in {{{1}}} for {{{n}}} and {{{2}}} for the value:

    {{{a(1)^2 + b(1) + c = a + b + c = 2}}}

The second term (that is, the term when {{{n = 2}}}) is 5, so:

    {{{a(2)^2 + b(2) + c = 4a + 2b + c = 5}}}

The third term (that is, the term when {{{n = 3}}}) is {{{10}}}, so:

    {{{a(3)^2 + b(3) + c = 9a + 3b + c = 10}}}


{{{a + b + c = 2}}}.........1
{{{4a + 2b + c = 5}}}...........2
{{{9a + 3b + c = 10}}}...........3
___________________________
{{{a + b + c = 2}}}.........1

{{{a  = 2- b -c}}}...plug in 2..........................a

{{{4(2- b -c)+ 2b + c = 5}}}...........2
{{{8- 4b -4c+ 2b + c = 5}}}
{{{8- 2b -3c= 5}}}
{{{8- 5-3c= 2b}}} 
{{{3-3c= 2b}}} 
{{{3/2-3c/2= b}}} ...........................b

{{{9(2- b -c) + 3b + c = 10}}}...........3...plug in a

{{{18-9 b -9c+ 3b + c = 10}}}

{{{18-6 b -8c  = 10}}}......plug in b

{{{18-6(3/2-3c/2)-8c = 10-18}}}
{{{-9+9c -8c = -8}}}
{{{c  = -8+9}}}
{{{c  =1}}}

find {{{a}}} and {{{b}}}:

{{{3/2-3c/2= b }}}
{{{3/2-3*1/2= b }}}
{{{highlight(b=0 )}}}
{{{a  = 2- b -c}}}
{{{a  = 2- 0 -1}}}
{{{highlight(a=1)}}}


so, we have {{{a = 1}}}, {{{b = 0}}}, and {{{c = 1}}}, and the formula is:

    1n^2 + 0n + 1 = {{{n^2 + 1}}}



so, a sub n ={{{n^2+1}}}(for all terms given)

3. -2,-8,-18,-32,-50,...a sub n =



-2 ----  -8 ----  -18----   -32 ----  -50

----..-6----..-10----..-14---..-18

----       .. -4----   .. -4----  ..  -4.........third difference, -> 3rd degree; constant is -4, so half of it (or -2)will be coefficient of n^2 

a sub n = {{{-2n^2 }}}(for all terms given)