Expansion of (1+2)nI was asked to show that ∀n∈N there exist a p∈N⋅ such that (1+2)n=p+p−1I used induction but it wasn't fruitful, so I tried to use the binomial expansion of (1+2)n but it seems I lack some insight to go further. Any hint is welcomed.

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Expansion of (1+2)nI was asked to show that ∀n∈N  there exist a p∈N⋅ such that (1+2)n=p+p−1I used induction but it wasn&#039;t fruitful, so I tried to use the binomial expansion of (1+2)n  but it seems I lack some insight to go further. Any hint is welcomed.

MoxboasteBots5h · Accepted Answer

The binomial formula shows you that  (1+2)n=an+bn2  for some integers an,bn.  But, the same binomial formula shows you that (convince yourself of this)  (1−2)n=an−bn2  for the same integers an,bn.  Then comes the hint: Calculate both  (an+bn2)(an−bn2)  and  (1+2)n(1−2)n=[(1+2)(1−2)]n  and compare.  You will get that a2n−2b2n=(−1)n, so a2n and 2b2n  differ from each other by one, and p will be the larger of the two.

Mary Goodson · Accepted Answer

This is a bit more roundabout than Jyrkis

Vasquez · Accepted Answer

It&#039;s reduced to show that the following expression––––––––––––––––––––– is an integer: [(1+2)−n2]2=14[(1+2)2n+(1−2)2n]+1212+14[(1+2)2n+(1−2)2n]=12+14[∑k=02n((2n),(k))2k2+∑k=02n((2n),(k))(−1)k2k2]=12+14[2∑k=0n((2n),(2k))2k]=12+12[1+∑k=1n((2n),(2k))2k]=1+∑k=1n((2n),(2k))2k−1

Expansion of (1+\sqrt2)^n I was asked to show that

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