Easy Complex question (part-2)

$Weight ~used ~are ~1, 3.~~~ .........1,~~ 3-1=2,~~ 3,~~ 1+3=4.\\ To~ get~ 5,~ we~ should~ have~ X,~ so~ X-4=5.~ So~ X=9.~ With~~{\color{#20A900}{1,~3,~9.}}......we~get~~9-4=5,~~9-3=6,~~9+1-3=7,~~9-1=8,~~9, \\ ~~9+1=10,~~9+3 - 1=11,~~9+3=12,~~9+3+1=13.\\ To~ get ~14,~ we~ should~ have~ Y,~ so~ Y-13=14.~~So~Y=27.~~~~~With~{\color{#EC7300}{1,~3,~9,~27}}~ we ~can ~get~ up~ to~ 13+27=40~~as~ above.\\ To~ get ~41,~ we~ should~ have~ Z,~ so~ Z-40=41.~~So~Z=81.~~~~~With~{\color{#3D99F6}{1,~3,~9,~27,~81}~} we ~can ~get~ up~ to~ 40+81=121~~as~ above.\\ So~there~are~only~{\color{#D61F06}{FIVE}}~weights~required.$

Each weight may be used in three different ways. It may be put on the left side, the right side, or not used at all. Therefore, if we have n weights, they may be combined in $3^{n}$ ways. This is true because in adding the weights, there are three choices for the coefficient of each weight in the sum: a plus sign if it’s on the left, a minus sign if it’s on the right, and a zero if it’s not used at all. There are, however, duplicates among these $3^{n}$ combinations. For every positive number, there is its negative (where the left and right scales are simply reversed), which represents the same weight on the balance scale. Since it is, in principle, possible for the number 0 to not be repeated, an upper bound on the number of positive integer weights that may be weighed with n fixed weights is (3^n - 1)/2. Therefore, to weigh all weights up to 121, we must have n ≥ 5.

The weights should be 1, 3, 9, 27, 81