Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We need to prove \[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ N and \[0 < x < \frac{\pi}{2}\] using mathematical induction.
For n = 1,
LHS = \[\frac{1}{2}\tan\frac{x}{2}\] and
\[RHS = \frac{1}{2}\cot\frac{x}{2} - \cot x = \frac{1}{2\tan\frac{x}{2}} - \frac{1}{\tan x}\]
\[ \Rightarrow RHS = \frac{1}{2\tan\frac{x}{2}} - \frac{1}{\frac{2\tan\frac{x}{2}}{1 - \tan^2 \frac{x}{2}}}\]
\[ \Rightarrow RHS = \frac{1}{2\tan\frac{x}{2}} - \frac{1 - \tan^2 \frac{x}{2}}{2 \tan\frac{x}{2}} = \frac{1 - 1 + \tan^2 \frac{x}{2}}{2\tan\frac{x}{2}} = \frac{\tan\frac{x}{2}}{2}\]
Therefore, the given relation is true for n = 1.
Now, let the given relation be true for n = k.
We need to prove that the given relation is true for n = k + 1.
\[\Rightarrow L = \frac{1}{2^k}\cot\left( \frac{x}{2^k} \right) + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
\[ \Rightarrow L = \frac{1}{2^k \tan\left( \frac{x}{2^k} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
\[ \Rightarrow L = \frac{1}{2^k \tan2\left( \frac{x}{2^{k + 1}} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
\[ \Rightarrow L = \frac{1}{2^k \times \frac{2\tan\left( \frac{x}{2^{k + 1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right)}} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
\[ \Rightarrow L = \frac{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right)}{2^{k + 1} \tan\left( \frac{x}{2^{k + 1}} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
\[\Rightarrow L = \frac{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right) + \tan^2 \left( \frac{x}{2^{k + 1}} \right)}{2^{k + 1} \tan\left( \frac{x}{2^{k + 1}} \right)} - \cot x = \frac{1}{2^{k + 1}}\cot\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
Now,
\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^k}\tan\left( \frac{x}{2^k} \right) + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) = \frac{1}{2^{k + 1}}\cot\left( \frac{x}{2^{k + 1}} \right) - \cot x\]
Thus,
\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ N and \[0 < x < \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
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:
`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`
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: 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.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is 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.
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
72n + 23n−3. 3n−1 is divisible by 25 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 the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
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:
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
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
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.
Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.
Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.
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))`
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
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 ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer
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:
4n – 1 is divisible by 3, for each natural number n.
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 that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
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.
