Question

Show that the highest power of n which is contained in $ {n^r} - 1 $ is equal to $ \dfrac{{{n^r} - nr + r - 1}}{{n - 1}} $

Answer 1

Hint: o find the highest power p of the number n where n is any positive integer use Legendre’s formula. Substitute $ {n^r} - 1 $ and n appropriately in the below formula.
Formula used: Legendre’s formula $ {h_p}\left( f \right) = \sum\limits_{i = 1}^\infty {\left\lfloor {\dfrac{f}{{{p^i}}}} \right\rfloor } $ where $ {h_p}\left( f \right) $ is the exponent of the largest power of n which uses the floor function, f is $ {n^r} - 1 $ which should be a positive integer and p is n and i vary from 1 to infinity.

Complete step-by-step answer:
We are given that to find the highest power of n in $ {n^r} - 1 $ and to prove the values is $ \dfrac{{{n^r} - nr + r - 1}}{{n - 1}} $
Using Legendre’s formula, we can find the exponent of the highest power of n contained in $ {n^r} - 1 $
 $
  {h_p}\left( f \right) = \sum\limits_{i = 1}^\infty {\left\lfloor {\dfrac{f}{{{p^i}}}} \right\rfloor } \\
  p = n,f = {n^r} - 1 \\
  {h_n}\left( {{n^r} - 1} \right) = \sum\limits_{i = 1}^\infty {\left\lfloor {\dfrac{{{n^r} - 1}}{{{n^i}}}} \right\rfloor } \\
   = \left\lfloor {\dfrac{{{n^r} - 1}}{{{n^1}}}} \right\rfloor + \left\lfloor {\dfrac{{{n^r} - 1}}{{{n^2}}}} \right\rfloor + \left\lfloor {\dfrac{{{n^r} - 1}}{{{n^3}}}} \right\rfloor + ....... + \left\lfloor {\dfrac{{{n^r} - 1}}{{{n^r}}}} \right\rfloor \\
  $
i must be from 1 to r, because when i value exceeds r, the exponent values of n will be less than 1.
So i is from 1 to r.
 $
   = \sum\limits_{k = 1}^r {\left\lfloor {\dfrac{{{n^r} - 1}}{{{n^k}}}} \right\rfloor } ,r \geqslant k \\
   = \sum\limits_{k = 1}^r {\left\lfloor {\dfrac{{{n^r}}}{{{n^k}}} - \dfrac{1}{{{n^k}}}} \right\rfloor } \\
   = \sum\limits_{k = 1}^r {\left\lfloor {{n^{r - k}} - \dfrac{1}{{{n^k}}}} \right\rfloor } \\
  \left( {\because \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}},m \geqslant n} \right) \\
   = \sum\limits_{k = 1}^r {\left\lfloor {{n^{r - k}}} \right\rfloor } + \sum\limits_{k = 1}^r {\left\lfloor { - \dfrac{1}{{{n^k}}}} \right\rfloor } \\
   = {n^{r - 1}} + {n^{r - 2}} + {n^{r - 3}} + ... + {n^{r - r}} - \dfrac{1}{{{n^1}}} - \dfrac{1}{{{n^2}}} - \dfrac{1}{{{n^3}}} - ... - \dfrac{1}{{{n^r}}} \\
   = \dfrac{{{n^r} - n}}{{n - 1}} - \left( {r - 1} \right) \\
   = \dfrac{{{n^r} - n - \left( {r - 1} \right)\left( {n - 1} \right)}}{{n - 1}} \\
   = \dfrac{{{n^r} - n - nr + r + n - 1}}{{n - 1}} \\
   = \dfrac{{{n^r} - nr + r - 1}}{{n - 1}} \\
  $
Therefore the highest power of n in $ {n^r} - 1 $ is $ \dfrac{{{n^r} - nr + r - 1}}{{n - 1}} $
 Hence, proved.

Note: An exponent is a number or letter written above and the right superscript of a mathematical expression called the base. It indicates that the base is to be raised to a certain power. The exponent corresponds to the number of times the base is used as a factor. Power denotes the repeated multiplication of a factor and the number which is raised to that base factor is the exponent. If $ {x^n} $ is a power then x is the base and n is the exponent. Do not confuse power with exponent.