Step-by-step explanation:
n² + n < 2^n
so, let's see what happens, when I increase n by 1
(n+1)² + n + 1 < 2^(n+1) = 2×2^n
n² + 2n + 1 + n + 1 < 2×2^n
n² + 3n + 2 < 2×2^n = 2^n + 2^n
so, we added on the left and on the right side of the inequality sign compared to the original relation :
2n + 2 vs. 2^n
so, once the original relation becomes true (and we see by trying that for n = 1, 2, 3 and 4 it is not, but for n = 5 it becomes true), the next integer follows the original inequality plus the expressions mentioned above.
n² + n + (2n + 2) < 2^n + (2^n)
this inequality will remain true, if
2n + 2 <= 2^n
n + 1 <= 2^(n-1), which is true for n > 2 (and therefore also n > 4), because the "power of" function 2^n progresses much faster than a simple n+1
and therefore this relation remains true for any larger n.
