Advertisements
Advertisements
Question
Solution
\[\text { Let } P\left( n \right): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 . \]
\[\text{ Step I: For } n = 2, \]
\[P\left( 2 \right): \]
\[LHS = \sqrt{2} \approx 1 . 414\]
\[RHS = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{\sqrt{2}}{2} \approx 1 + 0 . 707 = 1 . 707\]
\[\text{ As, LHS < RHS } \]
\[\text{ So, it is true for } n = 2 . \]
\[ \text{Step II : For } n = k \]
\[\text{ Let} P\left( k \right): \sqrt{k} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{k}} \text{ be true for some natural numbers } n \geq 2 . \]
\[\text{ Step III: For } n = k + 1, \]
\[P\left( k + 1 \right): \]
\[LHS = \sqrt{k + 1}\]
\[RHS = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}}\]
\[ > \sqrt{k} + \frac{1}{\sqrt{k + 1}}\]
\[\text{ As, } \sqrt{k + 1} > \sqrt{k}\]
\[ \Rightarrow \frac{\sqrt{k}}{\sqrt{k + 1}} < 1\]
\[ \Rightarrow \frac{k}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \frac{k + 1}{\sqrt{k + 1}} - \frac{1}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \sqrt{k + 1} - \frac{1}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \sqrt{k} + \frac{1}{\sqrt{k + 1}} > \sqrt{k + 1}\]
\[i . e . LHS < RHS\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence,} P\left( n \right): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]
APPEARS IN
RELATED QUESTIONS
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:
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: x2n – y2n is divisible by x + y.
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 "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
Given an example of a statement P (n) such that it is true for all n ∈ N.
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.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
52n+2 −24n −25 is divisible by 576 for all n ∈ 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?
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
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:
3 + 7 + 11 + ..... + to n terms = n(2n+1)
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1
Prove by method of induction, for all n ∈ N:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
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))`
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.
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`
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:
`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.
