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

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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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