# Prove that `(2*34^n )-( 3* 23^n) +1` is divisible by 726 for all positive integers n.

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Show that `2*34^n-3*23^n+1` is divisible by 726 for all positive integer n:

Rewrite as `2[(3(11)+1)^n]-3[(2(11)+1)^n]+1`

Expand each binomial:

`2[3^n11^n+_nC_(n-1)3^(n-1)11^(n-1)+ * * * _nC_1 3*11+1]` -

`3[2^n11^n+_nC_(n-1) 2^(n-1)11^(n-1)+ * * * _nC_1 2(11)+1]+1`

For n>1, every expression in the first bracket consists of all terms with factors of 3 and `11^2` , and every expression in the second bracket consists of all terms with factors of 2 and `11^2 ` except the last two terms in the brackets. So we rewrite:

`2[3(11^2)k+n(3*11)+1]-3[2(11^2)m+n(2*11)+1]+1` where k,m are integers.

Thus we have:

`726k+66n+2-726m-66n-3+1`

`=726(k-m)`

**So the expression can be written as a product of 726 and an integer, thus it is divisible by 726.**

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Ex: If n=4:

`2[3^4 11^4+4*3^3 11^3+6*3^2 11^2+4*3*11+1]-`

`3[2^4 11^4+4*2^3 11^3+6*2^2 11^2+4*2*11+1]+1`

`=162*11^4+216*11^3+108*11^2+24*11+2-48*11^4-96*11^3-72*11^2-24*11-3+1` `=114*11^4+120*11^3+36*11^2`

`=726(19*11^2+20*11+6)`

` ` `=726(2525)`

Or in terms of the proof, 726[3681-1156] where k=3681 and m=1156.

*** For n=1 we get 2(34)-3(23)+1=0 = 726*0. ***

**Sources:**