Question

Use mathematical induction to prove n! ≥ 2 n−1 for all integers n ≥ 1.

Answer 1

Answer:

Yes of course it's correct

Answer 2

Tip:

If $P(n)$ is a statement involving natural number $n$, such that

(i) $P(1)$ is true, i.e., the statement is true for $n=1$

(ii) If $P(k+1)$ is true whenever $P(k)$ is true , then $P(n)$ is true for all natural numbers.

Step

Step 1 of 3:

Let $P(n):n!\geq 2n-1;n\geq1$                         (1)

When $n=1$,     $P(1):1!\geq2(1)-1\\~~~~~~~~~~1\geq1$           for $n\geq1$, which is true

∴ $P(1)$ is true.

Step 2 of 3:

Let us assume that $P(k)$ is true for any positive integer $k$

     $P(k):k!\geq2k-1, ~n\geq1$                                (2)

Step 3 of 3:

Now, we shall prove that $P(k+1)$ is also true

    $P(k+1):(k+1)!\geq2(k+1)-1,~n\geq1$            (3)  

Now, $~~~~~~~~k!\geq2k-1;n\geq1$      

Multiplying both sides by $(k+1)$

$(k+1)k!~\geq(k+1)(2k-1);n\geq1\\(k+1)!\geq2k^2-k+2k-1\\(k+1)!\geq2k^2+k-1$                           (4)

Here, $k\in N$ so, $k^2\geq1$

$2k^2+k-1\geq2k+1$

So, from (4) we have, $(k+1)!\geq 2k+1$

Thus,$P(k+1)$ is true, whenever  $P(k)$ is true .  

Final Answer:

By mathematical induction, $P(n)$ is true for all integers $n\geq1$.