Question #11324

prove by mathematical induction that 2 times 34^n - 3 times 23^n + 1 is divisible by 726 for all positive integers n.

Expert's answer

Let S(n)=2*34^n-3*23^n+1

For n=1: S(1)=0 is divisible by 726

For n=2: S(2)=726 is divisible by 726

Let us assume S(n) is divisible by 726

S(n+1)=2*34^(n+1)-3*23^(n+1)+1=2*34*34^n-3*23*23^n+1=23(2*34^n-3*23^n)+11*2*34^n+1=23(2*34^n-3*23^n+1)+ +(11*2*34^n-22)=23S(n)+22(34^n-1)=

23S(n)+22*33(34^(n-1)+...+1)=23S(n)+726(34^(n-1)+...+1) is divisible by 726

according to assumption

So by mathematical induction 2*34^n-3*23^n+1 is divisible by 726 for all positive integers n.

For n=1: S(1)=0 is divisible by 726

For n=2: S(2)=726 is divisible by 726

Let us assume S(n) is divisible by 726

S(n+1)=2*34^(n+1)-3*23^(n+1)+1=2*34*34^n-3*23*23^n+1=23(2*34^n-3*23^n)+11*2*34^n+1=23(2*34^n-3*23^n+1)+ +(11*2*34^n-22)=23S(n)+22(34^n-1)=

23S(n)+22*33(34^(n-1)+...+1)=23S(n)+726(34^(n-1)+...+1) is divisible by 726

according to assumption

So by mathematical induction 2*34^n-3*23^n+1 is divisible by 726 for all positive integers n.

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