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: 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: `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:
Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
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 "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
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 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
32n+7 is divisible by 8 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 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?
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
Prove 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:
1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`
Prove by method of induction, for all n ∈ N:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
Prove by method of induction, for all n ∈ N:
`1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`
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
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2.
State whether the following proof (by mathematical induction) is true or false for the statement.
P(n): 12 + 22 + ... + n2 = `(n(n + 1) (2n + 1))/6`
Proof By the Principle of Mathematical induction, P(n) is true for n = 1,
12 = 1 = `(1(1 + 1)(2*1 + 1))/6`. Again for some k ≥ 1, k2 = `(k(k + 1)(2k + 1))/6`. Now we prove that
(k + 1)2 = `((k + 1)((k + 1) + 1)(2(k + 1) + 1))/6`
Prove the statement by using the Principle of Mathematical Induction:
4n – 1 is divisible by 3, for each natural number n.
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, 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.
Prove the statement by using the Principle of Mathematical Induction:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
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.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
