সমাধান:

 বামপক্ষ `= (4^n - 1)/(2^n - 1)`

          `= ((2^2)^n - 1)/(2^n - 1)`

         `= ((2^n)^2 - 1)/(2^n - 1)`

         `= ((2^n + 1)(2^n - 1))/((2^n - 1))` 
                              [ `:. a^2 - b^2 = (a + b)(a - b)` ]

         `= 2^n + 1`

           = ডানপক্ষ

    `:. (4^n - 1)/(2^n - 1) = 2^n + 1`  (প্রমাণিত)

সমাধান: 
 বামপক্ষ `= (2^(p + 1) .3^(2p - q) .5^(p + q) .6^q)/(6^p .10^(q + 2) .15^p)`

       `= (2^(p + 1) .3^(2p - q) .5^(p + q) .(2 xx 3)^q)/((2 xx 3)^p .(5 xx 2)^(q + 2) .(3 xx 5)^q)`

       `= (2^(p + 1) .3^(2p - q) .5^(p + q) .2^q .3^q)/(2^p .3^p .5^(q + 2) .2^(q + 2) .3^p .5^p)`

       `= (2^(p + q + 1) .3^(2p - q - q) .5^(p + q))/(2^(p + q + 2) .3^(p + p) .5^(q + p + 2))`
                    
       `= (2^(p + q + 1) .3^(2p) .5^(p + q))/(2^(p + q + 2) .3^(2p) .5^(p + q + 2))`

       `= 2^((p + q + 1) - (p + q + 2)) .3^(2p - 2p) .5^((p + q) - (p + q + 2))`

       `= 2^(p + q + 1 - p - q - 2) .3^0 .5^(p + q - p - q - 2)`
 
       `= 2^(- 1) . 1 . 5^(- 2)`
 
       `= 1/2 . 1 . 1/5^2`
    
       `= 1/2 . 1 . 1/25`

       `= 1/50`

  `:. (2^(p + 1) .3^(2p - q) .5^(p + q) .6^q)/(6^p .10^(q + 2) .15^p) `

      `= 1/50`   প্রমাণিত