SOLUTION: R0: 3,6,9,12,15,18,21,... R1: 1,4,7,10,13,16,19,... R2: 2,5,8,11,14,17,20,... a) Write down an expression for the value of the general term in each of the three prgressions. Sho

Algebra ->  Sequences-and-series -> SOLUTION: R0: 3,6,9,12,15,18,21,... R1: 1,4,7,10,13,16,19,... R2: 2,5,8,11,14,17,20,... a) Write down an expression for the value of the general term in each of the three prgressions. Sho      Log On


   



Question 1076171: R0: 3,6,9,12,15,18,21,...
R1: 1,4,7,10,13,16,19,...
R2: 2,5,8,11,14,17,20,...
a) Write down an expression for the value of the general term in each of the three prgressions. Show that the integer 1706836 lies in R1. I got R0:3n, R1:3n-2 and R2:3n-1. Then i did 3n-2=1706836 and showed that n=568946, so it lies in R1 and is the 568946th term in the series.
b) Use the Binomial Theorem to show that if x is a term in R1 or R2 then x^6 is in R1. This is the part i don't know how to work out.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a) I agree with your expressions for the the general term in each of the three sequences.
To show if any number belongs to one of them,
I would just add digits, using module 9 clock arithmetic. The number belongs in the sequence that contains the final sum.
If the ending sum is 0, 3, or 6, the number is divisible by 3, and belongs in R0.
If it is 1, 4, or 7, it belongs in R1; if 2,5, or 8, it belongs in R2.
For 1706836 ,
the final 36 part adds up to 3%2B69 , and that 9 is dropped.
The 1 and 8 add up to 1%2B8=9 , and that 9 is dropped.
Then we are left with 7%2B0%2B6=13 , and since 1%2B3=4 , we conclude that
when 1706836 is divided by 9 , the remainder is 4 .
So, 1706836=9K%2B4=3%2A3K%2B6-2=3%283K%2B2%29-2 for some integer K ,
and that shows that it belongs in R1.

HOW AND WHY CLOCK ARITHMETIC MOD 9 WORKS:
The sum of the digits of a number divisible by 9 is 9 or a multiple of 9.
As you are adding,
you drop partial sums that are 9, and keep adding the rest of the digits.
Similarly, for any partial two-digit sum,
you replace the sum of the digits for the two-digit partial sum.
At the end, you should have a one-digit final sum from 0 to 8.

b)The numbers in R1 and R2 are 1 away from a multiple of 3 .
We could write them all as 3p+%2B-+1 for some non-negative integer p
.
(If you - or your teacher - do not like that expression, you could use
3p+%2B+1 for terms of R1, with p=n-1 ,
and 3n+-+1 for terms of R2,
and write twice as much for your proof).
For x=3p+%2B-+1 ,
x%5E6=%283p+%2B-+1%29%5E6=%283p%29%5E6+%2B-+6%2A%281%29%283p%29%5E5%22%2B%2215%283p%29%5E4%281%29%5E2+%2B-+20%283p%29%5E3%281%29%5E3%22%2B%2215%283p%29%5E2%281%29%5E4+%2B-+6%283p%29%281%29%5E5%2B1%5E6
%22=%22%283p%29%5E6+%2B-+6%283p%29%5E5%22%2B%2215%283p%29%5E4+%2B-+20%283p%29%5E3%22%2B%2215%283p%29%5E2+%2B-+6%283p%29+ %22%2B+1%22
It should be obvious that all terms of that sum are multiples of 3 ,
except that %22+%2B+1%22 at the end,
and that shows that x%5E6 is a term in R1.

If your teacher insisted on seeing the n that gives x%5E6 the right to belong to R1,
you would have to write more, starting with getting out 3 as a common factor:
x%5E6+=+3%22%5B%22p%2A%283p%29%5E5+%2B-+2%283p%29%5E5%22%2B%225%283p%29%5E4+%2B-+20p%283p%29%5E2%22%2B%225%283p%29%5E2%281%29%5E4+%2B-+2%283p%29%22%5D%22 %22%2B+1%22 .