Question

Help!

Answer 1

\[\LARGE (1+x)(1+x^2)(1+x^4)...(1+x^{128})=\sum_{r=0}^{n} x^r\]
Solve for n.

Answer 2

Attempt..multiplying by (1-x) on both sides..
\[\LARGE (1-x^{256})=(1-x)(1+x+x^2+x^3...+x^n)\]

Answer 3

\[\LARGE a^n-b^n=(a-b)(a^{n-2}+a^{n-1}b+.....b^{n-1})\]

Answer 4

just an initial thought
wouldn't n be the highest power of x on left side?
x*x^2*x^4...x^128 = x^n

n = 1+2+4+....128
sum of geometric series

Answer 5

didn't get it :o

Answer 6

from your method you are getting n=255

from my method n = 1+2+4+....128 = 255 also

Answer 7

how do i compare "n" from my method?n i didn't get ur method

Answer 8

so you got stuck at very last step in your method :P

\(\LARGE a^N-b^N=(a-b)(a^{N-2}+a^{N-1}b+.....b^{N-1}) \\ \Large 1^N-x^N = ....\)

N = 256
N-1 = 255

Answer 9

notice your last term in 2nd bracket = x^n
and using the formula its x^(N-1)

thats why n = N-1 = 256-1 = 255

Answer 10

where did the minus go?? its +b^N-1 and -x^n ?

Answer 11

there's +x^n
not -x^n

Answer 12

its (1−x^256) on lhs O.o

Answer 13

so there's -b^n on lhs too...
why u getting confused with signs??
signs exactly match with the standard formula

Answer 14

draw karke batau kya?

Answer 15

yes !