Find the sum of all natural numbers n, satisfying the condition: divides
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Let's call 1 1 n + 2 n + 1 = F ( n ) .For some n, let F ( n ) divide F ( n + 1 ) .Thus, F ( n ) ∣ F ( n + 1 ) − 1 1 . F ( n ) ⇒ F ( n ) ∣ − 9 . 2 n − 1 0 and that means that F ( n ) ∣ 9 . 2 n + 1 0 ⇒ F ( n ) ≤ 9 . 2 n + 1 0 .So we get 1 1 n ≤ 8 . 2 n + 9 .But we can easily prove by induction that for every natural number n>1, 1 1 n > 8 . 2 n + 9 .Then we are left with only n=1 and we see that it satisfies the condition so the answer is 1