Question

Prove that 7 divides 2^n+1 + 3^2n-1 whenever n is a positive integer. on Promotion

Answer 1

Explanation:

To prove 7 divides
`2^(n+1) +3^(2n-1)`

let
`P\left [ n\right ]=2^(n+1) +3^(2n-1)`

for n>1

For n=1
`2^(1+1) +3^(2-1)=7 \left ( divisible by 7\right )`

to show that
`P\left ( n+1\right )=P\left ( n\right )`

`2^(n+1) +3^(2n-1)=7x` for some interger x

this equation will also be valid for
`P\left [ n+1\right ]=2^(\left ( n+1\right )+1)+3^(2\left ( n+1\right )-1)`

`P\left [ n+1\right ]=2^(\left ( n+1\right )+1)+3^(2n+1)`

rearranging

`P\left [ n+1\right ]=2\dot 2^(n+1) +3^2\dot 3^(2n-1)`

`P\left [ n+1\right ]=2\dot 2^(n+1) +9\dot 3^(2n-1)`

`P\left [ n+1\right ]=2\dot 2^(n+1) +\left ( 7+2\right )\dot 3^(2n-1)`

`P\left [ n+1\right ]=2\dot 2^(n+1) +\left ( 7\right )\dot 3^(2n-1)+\left ( 2\right )\dot 3^(2n-1)`

`P\left [ n+1\right ]=2\dot \left ( 2^(n+1) +3^(2n-1)\right )+\left ( 7\right )\dot 3^(2n-1)`

`P\left [ n+1\right ]=2\dot P\left ( k\right )+\left ( 7\right )\dot 3^(2n-1)`

using induction hyposthesis

`2^(n+1) +3^(2n-1)=7x`

thus

`P\left [ n+1\right ]=2\dot 7x +\left ( 7\right )\dot 3^(2n-1)`

`P\left [ n+1\right ]=7\left ( 2x+3^(2n-1)\right )`

hence 7 divides
`2^(n+1) +3^(2n-1)`