Question

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

Answer 1

Answer:

Step-by-step 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 $2^{n+1} +3^{2n-1}$