Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
Let P(n) be the given statement.
Thus, we have:
\[P\left( n \right): 1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\]
\[\text{ Step} 1: P(2): \frac{1}{2^2} = \frac{1}{4} < 2 - \frac{1}{2}\]
\[\text{ Thus, } P\left( 2 \right) \text{ is true } . \left[ \text{ We have not taken n = 1 because it is not possible . We will start this function from n = 2 onwards . } \right]\]
\[\text{ Step} 2: \]
\[\text{ Let P } \left( m \right) \text{ be true .} \]
\[\text{ Now } , \]
\[1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} < 2 - \frac{1}{m}\]
\[\text{ We need to prove that P(m + 1) is true } . \]
\[\text{ We know that P(m) is true } . \]
\[\text{ Thus, we have: } \]
\[1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} < 2 - \frac{1}{m}\]
\[ \Rightarrow 1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} + \frac{1}{\left( m + 1 \right)^2} < 2 - \frac{1}{m} + \frac{1}{\left( m + 1 \right)^2} \left[ \text{ Adding } \frac{1}{(m + 1 )^2} to \text{ both sides } \right]\]
\[ \Rightarrow P\left( m + 1 \right) < 2 - \frac{1}{m + 1} \left[ \because \left( m + 1 \right)^2 > m + 1, \frac{1}{\left( m + 1 \right)^2} < \frac{1}{m + 1} \Rightarrow \frac{1}{m} - \frac{1}{\left( m + 1 \right)^2} < \frac{1}{m + 1} as m < m + 1 \right] \]
\[\text{ Thus, } P\left( m + 1 \right) \text{ is true } . \]
\[\text{ By principle of mathematical induction, P(n) is true for all n } \in N, n \geq 2 .\]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`
Prove the following by using the principle of mathematical induction for all n ∈ N:
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
Given an example of a statement P (n) such that it is true for all n ∈ N.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
52n −1 is divisible by 24 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.
Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
Prove the statement by using the Principle of Mathematical Induction:
23n – 1 is divisible by 7, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
32n – 1 is divisible by 8, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Prove the statement by using the Principle of Mathematical Induction:
n3 – n is divisible by 6, for each natural number n ≥ 2.
Prove the statement by using the Principle of Mathematical Induction:
n(n2 + 5) is divisible by 6, for each natural number n.
A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
By using principle of mathematical induction for every natural number, (ab)n = ______.
