hmm. A re look at the question, I'm in a quandary. Depends on where the series ends and on the sign of x. When x>0 and when it ends with +2 it is x+2 When x>0 and when it ends with -2 it is x When x<0 and the series end with -2, it is |x|+2 when x<0 and the series ends with +2, it is |x|+4
That would be x.
ReplyDeletehmm. A re look at the question, I'm in a quandary.
ReplyDeleteDepends on where the series ends and on the sign of x.
When x>0 and when it ends with +2 it is x+2
When x>0 and when it ends with -2 it is x
When x<0 and the series end with -2, it is |x|+2
when x<0 and the series ends with +2, it is |x|+4
aah!Again a silly mistake .
ReplyDeleteWhen x>0 and when the series ends in +2 it is x
When x>0 and ends with -2, it's x-2
Superb effort implex. Keep 'em coming!
It can also have two additional values, 0 and 2 when x=2.
ReplyDeletex or x-2
ReplyDelete4-x ,2-x
@Riyaziq
ReplyDeleteAll the cases fall into only those four categories. Even for 2.
None of the solutions have been sastisfactory
ReplyDeleteLet |x-2|=y y>=0
after every even number of mods we will again get y ( as y+2 will be positive as y is so )
now if the series ends at +2
answer will be y+2 means |x-2|+2
if the series ends at -2
we will get y or |x-2|
The solution ends here, there is no need to do anything else
all the solutions will come in the ambit of this !
Cheers !
for x>=2 and
ReplyDeleteeven no. of mods...ans='x'
odd no. of mods...ans='x-2'
for x<2 and
even no. of mods...ans='2-x'
odd no. of mods...ans='4-x'
i think my soln...coincides with varun...[;)]
ReplyDelete