Question

Prove that \({n^2} + 1 \ge {2^n}\)when \(n\)is a positive integer with\(1 \le n \le 4\).

Short answer

\({n^2} + 1 \ge {2^n}\)is true for any positive integer n such that \(1 \le n \le 4\).

Step-by-step solution

Introduction

For \({n^2} + 1 \ge {2^n}\)whennis a positive integer with\(1 \le n \le 4\),can be shown by putting all the four values.

Case 1: For \(n = 1\)

Puttingn = 1,

1¹+=1+1

=2

≥2

=2¹

Case 2: For\(n = 2\)

Putting\(n = 2\),

\({2^2} + 1 = 4 + 1\)

\(\begin{array} = 5\\ > 4\\ = {2^2}\end{array}\)

Therefore for\(n = 2\), \({n^2} + 1 \ge {2^n}\)is true.

Case 3: For \(n = 3\)

Putting\(n = 3\),

\({3^2} + 1 = 9 + 1\)

\(\begin{array} = 10\\ > 8\\ = {2^3}\end{array}\)

Therefore, for\(n = 3\), \({n^2} + 1 \ge {2^n}\)is true.

Case 4: For \(n = 4\)

Putting\(n = 4\),

\({4^2} + 1 = 16 + 1\)

\(\begin{array} = 17\\ > 16\\ = {2^4}\end{array}\)

Therefore, for\(n = 4\),\({n^2} + 1 \ge {2^n}\)is true.

Hence, \({n^2} + 1 \ge {2^n}\)is true for any positive integer n such that \(1 \le n \le 4\).