Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let P(n): `sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, ∀ n ≥ 2
Step 1: P(2): `sqrt(2) < 1/sqrt(1) + 1/sqrt(2)` which is true.
Step 2: P(k): `sqrt(k) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)`
Let it be true.
Step 3: P(k + 1): `sqrt(k + 1) < 1/sqrt(2) + 1/sqrt(2) + ... + 1/sqrt(k + 1)`
From Step 2, we have
`sqrt(k) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)`
⇒ `sqrt(k) + 1/sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) + 1/sqrt(k + 1)`
⇒ `(sqrt(k) . sqrt(k + 1) + 1)/sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) + 1/sqrt(k + 1)` ......(i)
Now if `sqrt(k + 1) < (sqrt(k) . sqrt(k + 1) + 1)/sqrt(k + 1)`
⇒ `(k + 1) < sqrt(k) . sqrt(k + 1) + 1`
⇒ `k < sqrt(k) . sqrt(k + 1)` .......(ii)
From equation (i) and (ii) we get
`sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k + 1)`
Hence, P(k + 1) is true whenever P(k) is true.
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: 1.2.3 + 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 + 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:
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: 32n + 2 – 8n– 9 is divisible by 8.
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 "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
(ab)n = anbn for all n ∈ N.
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
Prove by method of induction, for all n ∈ N:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Answer the following:
Prove, by method of induction, for all n ∈ N
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Answer the following:
Prove by method of induction loga xn = n logax, x > 0, n ∈ N
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
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:
23n – 1 is divisible by 7, 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, 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:
n2 < 2n for all natural numbers n ≥ 5.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
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?
