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Challenging problem

Solution

Let’s start, we have
an+1=2n3an=2n3×2n1+32an1==2n+j=1n(3)j2nj+(3)n+1a0=2n+15+(a015)(3)n+1


If a0=15, then we have no problem. If a015, (23)n goes to 0 when n, so an3n will have the same sign as (a015)(1)n when n is large. Hence, an<an+1 will not hold then contradiction
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Challenging problem
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