Examples – Proof by Mathematical Induction for Inequalities

mathematical induction, prove inequalities using mathematical induction examples with solution, divisibility statements, summation identities

Prove Inequalities using Mathematical Induction examples with solutions

Question 1:

Prove the following inequality by the principle of mathematical induction:

{1^2} + {2^2} +  \ldots  + {n^2} > \frac{{{n^3}}}{3}, n N.

Solution:

We can write,

P\left( n \right):\quad {1^2} + {2^2} +  \ldots  + {n^2} > \frac{{{n^3}}}{3}

Step 1:

We first show that the basis for induction P(1) is true, that is, P(n) is true for n = 1.

Left Side = {1^2} = 1,

Right Side = \frac{{{1^3}}}{3} = \frac{1}{3}.

Since, 1 > \frac{1}{3}, that is, Left Side > Right Side

Hence, P(1) is true, that is, P(n) is true for n = 1.

Step 2:

Assume that P(k) is true for some natural number k, that is,

{1^2} + {2^2} +  \ldots  + {k^2} > \frac{{{k^3}}}{3}

We need to prove that P(k +1) is also true.

Because P(k) is true, so we have,

{1^2} + {2^2} +  \ldots  + {k^2} > \frac{{{k^3}}}{3}

Add {\left( {k + 1} \right)^2} to both sides:

{1^2} + {2^2} +  \ldots  + {k^2} + {\left( {k + 1} \right)^2}

> \frac{{{k^3}}}{3} + {\left( {k + 1} \right)^2}

= \frac{{{k^3} + 3{{\left( {k + 1} \right)}^2}}}{3}

= \frac{{{k^3} + 3\left( {{k^2} + 2k + 1} \right)}}{3}

= \frac{{{k^3} + 3{k^2} + 6k + 3}}{3}

= \frac{{{{\left( {k + 1} \right)}^3}}}{3}.

That is, we get

{1^2} + {2^2} +  \ldots  + {k^2} + {\left( {k + 1} \right)^2} > \frac{{{{\left( {k + 1} \right)}^3}}}{3}.

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

Therefore, from the principle of mathematical induction, the statement:

{1^2} + {2^2} +  \ldots  + {n^2} > \frac{{{n^3}}}{3}

is true for all natural number n.



Question 2:

Prove the following statement for all integers n \ge 4 by the principle of mathematical induction:

n! > {n^2}.

Solution:

We can write,

P\left( n \right):\quad n! > {n^2}

Step 1:

We first show that the basis for induction P(4) is true, that is, P(n) is true for n = 4.

Left Side = 4! = 4 \times 3 \times 2 \times 1 = 24,

Right Side = {4^2} = 16.

Since, 24 > 16, that is, 4! > {4^2}, that is, Left Side > Right Side

Hence, P(4) is true, that is, P(n) is true for n = 4.

Step 2:

Assume that P(k) is true for all integers k \ge 4, that is,

k! > {k^2}

We need to prove that P(k +1) is also true, that is, we need to prove that \left( {k + 1} \right)! > {\left( {k + 1} \right)^2} if k! > {k^2}.

To prove required from this, consider,

\left( {k + 1} \right)! = \left( {k + 1} \right) \cdot k!

> \left( {k + 1} \right) \cdot {k^2}                  (Apply k! > {k^2} by the induction hypothesis)

> 4{k^2}                               (Because k \ge 4 so k + 1 > 4

= {k^2} + 2{k^2} + {k^2}

> {k^2} + 2k + 1

= {\left( {k + 1} \right)^2}.

That is, we get

\left( {k + 1} \right)! > {\left( {k + 1} \right)^2}.

Thus, P(k +1) is true whenever P(k) is true for all integers k \ge 4.

Therefore, from the principle of mathematical induction, the statement:

n! > {n^2}

is true for all integers n \ge 4.


Tags: mathematical induction examples with solutions, prove inequalities by induction



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About Lata Agarwal 270 Articles
M.Phil in Mathematics, skilled in MS Office, MathType, Ti-83, Internet, etc., and Teaching with strong education professional. Passionate teacher and loves math. Worked as a Assistant Professor for BBA, BCA, BSC(CS & IT), BE, etc. Also, experienced SME (Mathematics) with a demonstrated history of working in the internet industry. Provide the well explained detailed solutions in step-by-step format for different branches of US mathematics textbooks.

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