Hence, [tex]n^2 + 1\geq 2^n[/tex] when [tex]n[/tex] is a positive integer with [tex]1 \leq n \leq 4[/tex].
Use principle of mathematical induction, put [tex]n=1,2,3,4[/tex] in the given inequality
For [tex]n=1[/tex]
[tex]1^2+1\geq2^1[/tex]
[tex]2\geq2[/tex], which is true.
So, [tex]n^2+1\geq2^n[/tex] is true for [tex]n=1[/tex]
For [tex]n=2[/tex]
[tex]2^2+1\geq2^2[/tex]
[tex]5\geq4[/tex], which is true.
So, [tex]n^2+1\geq2^n[/tex] is true for [tex]n=2[/tex]
For [tex]n=3[/tex]
[tex]3^2+1\geq2^3[/tex]
[tex]10\geq8[/tex], which is true.
So, [tex]n^2+1\geq2^n[/tex] is true for [tex]n=3[/tex]
For [tex]n=4[/tex]
[tex]4^2+1\geq2^4[/tex]
[tex]17\geq16[/tex], which is true.
So, [tex]n^2+1\geq2^n[/tex] is true for [tex]n=4[/tex]
Learn more about the proofs by principle of mathematical induction here:
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