Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): \frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n \text{ is a positive integer for all } n \in N . \]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{62}{165} = \frac{15 + 33 + 55 + 62}{165} = \frac{165}{165} = 1 \]
\[\text{ It is certainly a positive integer } . \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then, } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \text{ is a positive integer . } \]
\[\text{ Now, let } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m = \lambda, \text{ where } \lambda \in N\text{ is a positive integer . } \]
\[\text{ We have to show that P(m + 1) is true whenever P(m) is true } . \]
\[\text{ To prove: } \frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\text{ is a positive integer .} \]
\[\text{ Now, } \]
\[\frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\]
\[ = \frac{1}{11}\left( m^{11} + 11 m^{10} + 55 m^9 + 165 m^8 + 330 m^7 + 462 m^6 + 462 m^5 + 330 m^4 + 165 m^3 + 55 m^2 + 11m + 1 \right)\]
\[ + \frac{1}{5}\left( m^5 + 5 m^4 + 10 m^3 + 10 m^2 + 5m + 1 \right) + \frac{1}{3}\left( m^3 + 3 m^2 + 3m + 1 \right)\]
\[ + \frac{62}{165}m + \frac{62}{165}\]
\[ = \left[ \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \right] + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{6}{105}\]
\[ = \lambda + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + 1\]
\[\text{ It is a positive integer } . \]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/2`
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
Prove the following by using the principle of mathematical induction for all n ∈ N: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^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:
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: 41n – 14n is a multiple of 27.
Prove the following by using the principle of mathematical induction for all n ∈ N (2n +7) < (n + 3)2
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
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.
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
(ab)n = anbn for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ 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?
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
Prove by method of induction, for all n ∈ N:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ 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, 7n – 2n is divisible by 5.
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:
2n < (n + 2)! for all natural number 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.
A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.
Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
State whether the following statement is true or false. Justify.
Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
