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Tuesday, November 30, 2021

2021/104) Prove that for positive integer n we have n^2 | (n+1)^n-1

We have

(n+1)^n - 1

= \sum_{k=0}^{n}  {n \choose k} n^{n-k} -1

= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} + {n \choose n-1} n^{n-(n-1)} + {n \choose n} n^{n-n} -1

= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n * n + 1 - 1

= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n^2

now each term in sum is having n^2 as a term and hence the expression is divisible by n^2

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