Question

How do you find S(sub11) for1+2+4+8+...?

Answer 1

I think you are asking: "what is the sum of the first 11 terms". It is the 12th term minus 1.

Answer 2

It is simple like"1+1 = 2"
so it is not
except you add that the summoned may not be repeated

For example: N = { 1, 2, 3 ,...} is closed under addition
Z = { ... , -2, -1, 0 , 1, 2 , ...} is closed under add and sub.

Answer 3

hmmmm ok

Answer 4

Thank you :p

Answer 5

It's kinda like this "3 + 6 (mod 7)
= 9 (mod 7) [9 ÷ 7 = 1 (quotient) ... 2 (remainder)]
= 2 (mod 7)
"r (mod d) (r = remainder, d = divisor)"