Mathematics, 18.07.2019 20:20, 987654321156
Prove that 7 divides 2^n+1 + 3^2n-1 whenever n is a positive integer. on promotion
Answers: 2
![P\left [ n\right ]=2^{n+1} +3^{2n-1}](/tpl/images/0105/2416/61017.png)
for some interger x![P\left [ n+1\right ]=2^{\left ( n+1\right )+1}+3^{2\left ( n+1\right )-1}](/tpl/images/0105/2416/bb516.png)
![P\left [ n+1\right ]=2^{\left ( n+1\right )+1}+3^{2n+1}](/tpl/images/0105/2416/af0a4.png)
![P\left [ n+1\right ]=2\dot 2^{n+1} +3^2\dot 3^{2n-1}](/tpl/images/0105/2416/64208.png)
![P\left [ n+1\right ]=2\dot 2^{n+1} +9\dot 3^{2n-1}](/tpl/images/0105/2416/29bc8.png)
![P\left [ n+1\right ]=2\dot 2^{n+1} +\left ( 7+2\right )\dot 3^{2n-1}](/tpl/images/0105/2416/d4a3a.png)
![P\left [ n+1\right ]=2\dot 2^{n+1} +\left ( 7\right )\dot 3^{2n-1}+\left ( 2\right )\dot 3^{2n-1}](/tpl/images/0105/2416/b30e9.png)
![P\left [ n+1\right ]=2\dot \left ( 2^{n+1} +3^{2n-1}\right )+\left ( 7\right )\dot 3^{2n-1}](/tpl/images/0105/2416/f0328.png)
![P\left [ n+1\right ]=2\dot P\left ( k\right )+\left ( 7\right )\dot 3^{2n-1}](/tpl/images/0105/2416/e3e5d.png)
![P\left [ n+1\right ]=2\dot 7x +\left ( 7\right )\dot 3^{2n-1}](/tpl/images/0105/2416/7f723.png)
![P\left [ n+1\right ]=7\left ( 2x+3^{2n-1}\right )](/tpl/images/0105/2416/d29dd.png)
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Prove that 7 divides 2^n+1 + 3^2n-1 whenever n is a positive integer. on promotion...
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