Examples – Proof by Mathematical Induction for Summation Identities

Lata Agarwal Mathematical Induction, Mathpedia
mathematical induction, prove inequalities using mathematical induction examples with solution, divisibility statements, summation identities

Prove Summation Identities using Mathematical Induction examples with solutions

Question 1:

Prove the following statement by the principle of mathematical induction:

1 + 2 + 3 + \ldots + n = \frac{{n\left( {n + 1} \right)}}{2}, n ∈ N.

Solution:

We can write,

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

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,

Right Side = \frac{{1\left( {1 + 1} \right)}}{2} = \frac{{1\left( 2 \right)}}{2} = \frac{2}{2} = 1.

Both sides of the statement are equal for n = 1. 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 + 3 + \ldots + k = \frac{{k\left( {k + 1} \right)}}{2}

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

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

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

Add k + 1 to both sides:

1 + 2 + 3 + \ldots + k + \left( {k + 1} \right)

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

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

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

That is, we get

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

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

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

1 + 2 + 3 + \ldots + n = \frac{{n\left( {n + 1} \right)}}{2}

is true for all natural number n.

Question 2:

Prove the following statement by the principle of mathematical induction:

{1^2} + {2^2} + {3^2} + \ldots + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}, n ∈ N.

Solution:

We can write,

P\left( n \right):\quad {1^2} + {2^2} + {3^2} + \ldots + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}

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\left( {1 + 1} \right)\left( {2 \times 1 + 1} \right)}}{6} = \frac{{1\left( 2 \right)\left( 3 \right)}}{6} = \frac{6}{6} = 1.

Both sides of the statement are equal for n = 1. 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} + {3^2} + \ldots + {k^2} = \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{6}

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

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

{1^2} + {2^2} + {3^2} + \ldots + {k^2} = \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{6}

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

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

= \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{6} + {\left( {k + 1} \right)^2}

= \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right) + 6{{\left( {k + 1} \right)}^2}}}{6}

= \frac{{\left( {k + 1} \right)\left[ {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right]}}{6}

= \frac{{\left( {k + 1} \right)\left( {2{k^2} + k + 6k + 6} \right)}}{6}

= \frac{{\left( {k + 1} \right)\left( {2{k^2} + 7k + 6} \right)}}{6}

= \frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)}}{6}

= \frac{{\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)}}{6}.

That is, we get

{1^2} + {2^2} + {3^2} + \ldots + {k^2} + {\left( {k + 1} \right)^2} = \frac{{\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)}}{6}.

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

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

{1^2} + {2^2} + {3^2} + \ldots + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}

is true for all natural number n.

Tags: mathematical induction examples with solutions, prove summation identities by induction

Copyrighted Material © 2019 - 2024 Prinsli.com - All rights reserved

All content on this website is copyrighted. It is prohibited to copy, publish or distribute the content and images of this website through any website, book, newspaper, software, videos, YouTube Channel or any other medium without written permission. You are not authorized to alter, obscure or remove any proprietary information, copyright or logo from this Website in any way. If any of these rules are violated, it will be strongly protested and legal action will be taken.

About Lata Agarwal 268 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.
Website Facebook LinkedIn

Related Articles

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

Examples – Proof by Mathematical Induction for Divisibility Statements

Lata Agarwal Mathematical Induction, Mathpedia

Examples for Mathematical Induction Proofs for Divisibility Statements: Ex 1: Prove the following by the principle of mathematical… […]

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

Principle of Mathematical Induction

Lata Agarwal Mathematical Induction, Mathpedia

There are many mathematical results that can be proven using mathematical induction. For example, summation identities, divisibility statements, inequalities, etc. […]

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

What is mathematical induction with example?

Lata Agarwal Mathematical Induction, Mathpedia

In mathematics, mathematical induction is a very powerful tool. It is a technique for proving statements that hold to all natural numbers… […]

Be the first to comment

Leave a Reply

Your email address will not be published.


*