Advertisements
Advertisements
प्रश्न
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
Advertisements
उत्तर
Let P(n) be the given statement.
Thus, we have .
\[P\left( n \right): \frac{\left( 2n \right)!}{2^{2n} \left( n! \right)^2} \leq \frac{1}{\sqrt{3n + 1}}\]
\[\text{ Step1} : \]
\[P(1): \frac{2!}{2^2 . 1} = \frac{1}{2} \leq \frac{1}{\sqrt{3 + 1}}\]
\[\text{ Thus, P(1) is true} . \]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true .} \]
\[\text{ Thus, we have: } \]
\[\frac{\left( 2m \right)!}{2^{2m} \left( m! \right)^2} \leq \frac{1}{\sqrt{3m + 1}}\]
\[\text{ We need to prove that P(m + 1) is true .} \]
Now,
\[P(m + 1): \]
\[\frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} = \frac{\left( 2m + 2 \right)\left( 2m + 1 \right)\left( 2m \right)!}{2^{2m} . 2^2 \left( m + 1 \right)^2 \left( m! \right)^2}\]
\[ \Rightarrow \frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} \leq \frac{\left( 2m \right)!}{2^{2m} \left( m! \right)^2} \times \frac{\left( 2m + 2 \right)\left( 2m + 1 \right)}{2^2 \left( m + 1 \right)^2}\]
\[ \Rightarrow \frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} \leq \frac{2m + 1}{2\left( m + 1 \right)\sqrt{3m + 1}}\]
\[\Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{\left( 2m + 1 \right)^2}{4 \left( m + 1 \right)^2 \left( 3m + 1 \right)}}\]
\[ \Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{\left( 4 m^2 + 4m + 1 \right) \times \left( 3m + 4 \right)}{4\left( 3 m^3 + 7 m^2 + 5m + 1 \right)\left( 3m + 4 \right)}}\]
\[ \Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{12 m^3 + 28 m^2 + 19m + 4}{\left( 12 m^3 + 28 m^2 + 20m + 4 \right)\left( 3m + 4 \right)}}\]
\[ \because \frac{12 m^3 + 28 m^2 + 19m + 4}{\left( 12 m^3 + 28 m^2 + 20m + 4 \right)} < 1\]
\[ \therefore \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} < \frac{1}{\sqrt{3m + 4}}\]
Thus, P(m + 1) is true.
Hence, by mathematical induction
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] is true for all n ∈ N
APPEARS IN
संबंधित प्रश्न
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`
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
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.
(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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
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?
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
Prove by method of induction, for all n ∈ N:
3n − 2n − 1 is divisible by 4
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:
Prove, by method of induction, for all n ∈ N
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`
Answer the following:
Prove by method of induction
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
Answer the following:
Prove by method of induction loga xn = n logax, x > 0, n ∈ N
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
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, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.
Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`
A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.
Give an example of a statement P(n) which is true for all n. Justify your answer.
Prove the statement by using the Principle of Mathematical Induction:
n3 – 7n + 3 is divisible by 3, 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:
n3 – n is divisible by 6, for each natural number n ≥ 2.
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
