Question

prove that 64^(2n) -36^(2n) is divisible by 100​

Answer 1

Answer:

3^(2(k+1) + 2) - 8(k+1) - 9 - (3^(2k+2) - 8k - 9)

= 3^(2k + 4) - 8k - 8 - 9 - 3^(2k+2) + 8k + 9

= 3^(2k+4) - 3^(2k+2) - 8

= 3^(2k+2)( 3^2 - 1) - 8

= 3^(2k+2)(8) - 8

= 3^(2k)(3^2)(8) - 8

= 72(3^(2k)) - 8

= 72(3^2)^k - 8

= 72(9^k) - 8

= 8(9^(k+1)) - 8

= 8( 9^(k+1) - 1)

Step-by-step explanation:

Answer 2

Answer:

thank you for your free point