Show that `9^(n+1)-8n-9` is divisible by 64, where `n` is a positive integer.

Explanation

In order to show that `9^(n+1)-8n-9`is divisible by 64,

We have to prove that

` " " " "9^(n+1)-8n-9=64k`, where `k` is some natural number

Writing `9^(n+1)=(1+8)^(n+1)`

We know that

`(a+b)^(n)=^(n)C_0a^(n)+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

Putting `a=1`, `b=8` and `n=n+1`

`(9)^(n+1)=^(n+1)C_0(1)^(n+1)+^(n+1)C_1(1)^(n)(8)^1+^(n+1)C_2(1)^(n-1)(8)^2+.......+^(n+1)C_(n+1)(8)^(n+1)`

`(9)^(n+1)=^(n+1)C_0+^(n+1)C_1(8)+^(n+1)C_2(8)^2+.......+^(n+1)C_(n+1)(8)^(n+1)`

`(9)^(n+1)=1+((n+1)!)/(1!(n+1-1)!)(8)+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(1)(8)^(n+1)`

`(9)^(n+1)=1+((n+1)n!)/(n!)(8)+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

`(9)^(n+1)=1+(n+1)(8)+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

`(9)^(n+1)=1+(8n+8)+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

`(9)^(n+1)=8n+9+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

Hence,

`9^(n+1)=8n+9+^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

`9^(n+1)-8n-9=^(n+1)C_2(8)^2+^(n+1)C_3(8)^3+.......+(8)^(n+1)`

Taking `8^2` common from right side

`9^(n+1)-8n-9=(8)^2[ " ^(n+1)C_2+^(n+1)C_3(8)+.......+(8)^(n-1)]`

`9^(n+1)-8n-9=64[ " ^(n+1)C_2+^(n+1)C_3(8)+.......+(8)^(n-1)]`

`9^(n+1)-8n-9=64k`

` " " " "` where `k=[ " ^(n+1)C_2+^(n+1)C_3(8)+.......+(8)^(n-1)]` is a natural number

Thus, `9^(n+1)-8n-9`is divisible by 64

Hence Proved