1 + 2 + 3 + ... + n = n ( n + 1 ) 2 i.e. the sum of the first n natural numbers is n ( n + 1 ) 2 .

Question

1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .

Solution

Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n =\[\frac{n(n + 1)}{2}\]

\[\text{ Step} 1: \]

\[P(1) = 1 = \frac{1(1 + 1)}{2} = 1\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step } 2: \]

\[\text{ Let P(m) be true } . \]

\[\text{ Then, } \]

\[1 + 2 + 3 + . . . + m = \frac{m(m + 1)}{2}\]

\[\text{ We shall now prove that P(m + 1) is true } . \]

\[i . e . , \]

\[1 + 2 + . . . + (m + 1) = \frac{(m + 1)(m + 2)}{2}\]

\[\text{ Now,} \]

\[1 + 2 + . . . + m = \frac{m(m + 1)}{2}\]

\[ \Rightarrow 1 + 2 + . . . m + m + 1 = \frac{m(m + 1)}{2} + m + 1 \left[ \text{ Adding } \left( m + 1 \right) \text{ to both sides } \right]\]

\[ = \frac{m^2 + m + 2m + 2}{2}\]

\[ = \frac{(m + 1)(m + 2)}{2}\]

\[\text{ Hence, P(m + 1) is true } . \]

By the principle of mathematical induction, P(n) is true for all n \[\in\] N .

shaalaa.com

Principle of Mathematical Induction

Is there an error in this question or solution?

Q 1 Q 7 Q 2

Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11

Chapter 12 Mathematical Induction
Exercise 12.2 | Q 1 | Page 27

Video TutorialsVIEW ALL [1]

view
Video Tutorials For All Subjects
Principle of Mathematical Induction

video tutorial00:17:42

RELATED QUESTIONS

Prove the following by using the principle of mathematical induction for all n ∈ N: x²ⁿ – y²ⁿ is divisible by x + y.

Prove the following by using the principle of mathematical induction for all n ∈ N: 3²ⁿ^{+ 2} – 8n– 9 is divisible by 8.

If P (n) is the statement "n(n + 1) is even", then what is P(3)?

Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

1² + 2² + 3² + ... + n² =\[\frac{n(n + 1)(2n + 1)}{6}\] .

1 + 3 + 3² + ... + 3ⁿ⁻¹ = \[\frac{3^n - 1}{2}\]

1 + 3 + 5 + ... + (2n − 1) = n² i.e., the sum of first n odd natural numbers is n².

\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]

1² + 3² + 5² + ... + (2n − 1)² = \[\frac{1}{3}n(4 n^2 - 1)\]

5²ⁿ −1 is divisible by 24 for all n ∈ N.

(ab)ⁿ = aⁿbⁿ for all n ∈ N.

Given \[a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right)\] for n ≥ 2, where a > 0, A > 0.
Prove that \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]

\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

Show by the Principle of Mathematical induction that the sum S_n of then terms of the series \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

Prove that the number of subsets of a set containing n distinct elements is 2ⁿ, for all n \[\in\] N .

\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]

Prove by method of induction, for all n ∈ N:

3 + 7 + 11 + ..... + to n terms = n(2n+1)

Prove by method of induction, for all n ∈ N:

1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`

Prove by method of induction, for all n ∈ N:

(2³ⁿ − 1) is divisible by 7

Prove by method of induction, for all n ∈ N:

3ⁿ − 2n − 1 is divisible by 4

Answer the following:

Prove by method of induction 5²ⁿ − 2²ⁿ is divisible by 3, for all n ∈ N

Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

2²ⁿ – 1 is divisible by 3.

The distributive law from algebra says that for all real numbers c, a₁ and a₂, we have c(a₁ + a₂) = ca₁ + ca₂.

Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a₁, a₂, ..., a_n are any real numbers, then c(a₁ + a₂ + ... + a_n) = ca₁ + ca₂ + ... + ca_n.

Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.

Show by the Principle of Mathematical Induction that the sum S_n of the n term of the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² ... is given by

S_n = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`

Give an example of a statement P(n) which is true for all n. Justify your answer.