Advertisements
Advertisements
प्रश्न
n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): n(n + 1)(n + 5)\text{ is a multiple of } 3 . \]
\[\text{ Step1 } : \]
\[P(1): 1(1 + 1)(1 + 5) = 12 \]
\[\text{ It is a multiple of} 3 . \]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step2 } : \]
\[\text{ Let } P\left( m \right) \text{ be true . } \]
\[\text{ Then,} m\left( m + 1 \right)\left( m + 5 \right) \text{ is a multiple of } 3 . \]
\[\text{ Suppose} m\left( m + 1 \right)\left( m + 5 \right) = 3\lambda, \text{ where } \lambda \in N . \]
\[\text{ We have to show that } P\left( m + 1 \right) \text{ is true whenever P(m) is true } . \]
\[\text{ Now, } \]
\[P(m + 1) = \left( m + 1 \right)\left( m + 2 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 6 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 5 + 1 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 5 \right) + m(m + 1) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = 3\lambda + \left( m + 1 \right)\left( m + 2m + 6 \right) \left[ From P(m) \right]\]
\[ = 3\lambda + 3\left( m + 1 \right)\left( m + 2 \right)\]
\[\text{ It is clearly a multiple of 3} . \]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = `(n(n+1)(n+2)(n+3))/(4(n+3))`
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:
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:
(1+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N:
`(1+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`
Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
Given an example of a statement P (n) such that it is true for all n ∈ N.
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.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
52n −1 is divisible by 24 for all n ∈ N.
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]
\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]
Prove by method of induction, for all n ∈ N:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
Answer the following:
Prove, by method of induction, for all n ∈ N
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
Answer the following:
Prove by method of induction
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
Prove the statement by using the Principle of Mathematical Induction:
`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.
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.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
By using principle of mathematical induction for every natural number, (ab)n = ______.