Show that 9n+1 - 8n - 9 is divisible by 64 whenever n is a positive integer.

We have 9n+1 = (1 + 8)n+1 = ^ n + 1 C_0 + ^ n + 1 C_1 (8) + ^ n + 1 C_2 (8)^2 + ^ n + 1 C_3 (8)^3 + ... + ^ n + 1 C_ n + 1 (8)^ n + 1 = 1 + (n + 1) 8 + ^ n + 1 C_2 (8)^2 + ^ n + 1 C_3 (8)^3 + ... + ^ n + 1 C_ n + 1 (8)^ n + 1 = 1 + 8n + 8 + ^ n + 1 C_2 (8)^2 + ^ n + 1 C_3 (8)^3 + .... + ^ n + 1 C_ n + 1 (8)^ n + 1 font-family:Tahoma font-size:8px = ;9 ;+8n ;+^ n+1 C_2 (8 )^2 ;+^ n+1 C_3 (8 )^3 ;+........ ;+^ n+1 C_ n+1 (8)^ n+1 font-family:Tahoma font-size:8px 9^ n+1 -8n-9=^ n+1 C_2 (8)^2 +^ n+1 C_3 (8)^3 + ... + ^ n + 1 C_ n + 1 (8)^ n + 1 = 64 [^ n + 1 C_2 + ^ n + 1 C_3 8 + ... + ^ n + 1 C_ n + 1 8^ n + 1 ] which show that 9n+1 - 8n-9 is divisible by 64 wherever n is a positive integer

Show that 9n+1 - 8n - 9 is divisible by 64 whenever n is a positi…

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Question

Show that 9ⁿ⁺¹ - 8n - 9 is divisible by 64 whenever n is a positive integer.

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Answer

We have 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹
${ = ^{n + 1}}{C_0}{ + ^{n + 1}}{C_1}(8){ + ^{n + 1}}{C_2}{(8)^2} + $$^{n + 1}{C_3}{(8)^3} + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 1 + (n + 1) \times 8{ + ^{n + 1}}{C_2}{(8)^2}$${ + ^{n + 1}}{C_3}{(8)^3} + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 1 + 8n + 8{ + ^{n + 1}}{C_2}{(8)^2}{ + ^{n + 1}}{C_3}{(8)^3} + $$....{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$

$\style{font-family:Tahoma}{\style{font-size:8px}{=\;9\;+8n\;+^{n+1}C_2\left(8\right)^2\;+^{n+1}C_3\left(8\right)^3\;+........\;+^{n+1}C_{n+1}(8)^{n+1}}}$
$ \style{font-family:Tahoma}{\style{font-size:8px}{\Rightarrow9^{n+1}-8n-9=^{n+1}C_2{(8)^2}+^{n+1}C_3{(8)^3}}}$$ + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 64{[^{n + 1}}{C_2}{ + ^{n + 1}}{C_3} \cdot 8$$ + ...{ + ^{n + 1}}{C_{n + 1}} \cdot {8^{n + 1}}]$
which show that 9ⁿ⁺¹ - 8^n-9 is divisible by 64 wherever n is a positive integer