Question

Prove by induction that for all positive integers, 5^n>= n^2

Answer 1

so, assume that
\[5^k \ge k^2 \]
is true, if we can show that the statement is true for k+1, then the statement is true for all k
\[5^(k+1)\ge (k+1)^2\]
\[5^k * 5 \ge k^2 +2 k +1 \]

where do we go from here

Answer 2

\(5^k \ge k^2\)

multiply both sides by 5
\(5^{k+1} \ge 5k^2\)
\(~~~~~~~ \ge k^2 + 2k^2 + 2k^2\)
\(~~~~~~~ \ge k^2 + 2k + 1\)
\(~~~~~~~ \ge (k+1)^2\)

Answer 3

5^(k+1)=5^k *5= 5^k+4*5^k
4*5^k>4k^2 (as per assumption) > 2k^2+1

Answer 4

sorry, 4k^2>2k+1

Answer 5

combine those

Answer 6

if you don't like the earlier method,
you may use a sub proof to show 4k^2 > 2k+1

Answer 7

show that 4k^2 > 2k+1 before diving into induction step

Answer 8

that's easy, for integers, 4k^2>=4k>=3k+1>2k+1

Answer 9

for positive integers*

Answer 10

thank guys

Answer 11

yw !

Answer 12

So do we assume \(k>0\) then?

Answer 13

Since it is all positive integers, then \(n\geq 1\). So I guess it checks out.