Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{ Step I: For n } = 2, \]
\[LHS = \frac{1}{2 + 1} + \frac{1}{2 \times 2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} = \frac{14}{24} > \frac{13}{24} = RHS\]
\[As, LHS > RHS\]
\[\text{ So, it is true for n } = 2 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } p\left( k \right): \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} > \frac{13}{24}, \text{ be true for some natural numbers } k > 1 . \]
\[\text{ Step III: For n } = k + 1, \]
\[p\left( k + 1 \right) = \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} + \frac{1}{2k + 1}\]
\[ > \frac{13}{24} + \frac{1}{2k + 1} \left( \text{ Using step } II \right)\]
\[ > \frac{13}{24}\]
\[i . e . p\left( k + 1 \right) > \frac{13}{24}\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence,} p\left( n \right): \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers} n > 1 .\]
APPEARS IN
RELATED QUESTIONS
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:
Prove the following by using the principle of mathematical induction for all n ∈ N:
1.2 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/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: 102n – 1 + 1 is divisible by 11
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 "n(n + 1) is even", then what is P(3)?
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
Show by the Principle of Mathematical induction that the sum Sn of then terms of the series \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
Prove by method of induction, for all n ∈ N:
13 + 33 + 53 + .... to n terms = n2(2n2 − 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
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
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
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.
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 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, 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:
n(n2 + 5) is divisible by 6, for each natural number n.
Prove the statement by using the Principle of Mathematical Induction:
n2 < 2n for all natural numbers n ≥ 5.
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 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.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
