Advertisements
Advertisements
प्रश्न
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): x^{2n - 1} + y^{2n - 1} \text{ is divisible by } x + y . \]
\[\text{ Step1:} \]
\[P(1): x^{2 - 1} + y^{2 - 1} = x + y \text{ is divisible by } x + y\]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true } . \]
\[\text{ Also } , \]
\[ x^{2m - 1} + y^{2m - 1} \text{ is divisible by } x + y . \]
\[\text{ Suppose:} \]
\[ x^{2m - 1} + y^{2m - 1} = \lambda\left( x + y \right) \text{ where} \lambda \in N . . . (1)\]
\[\text{ We shall show that } P\left( m + 1 \right) \text{ is true whenever } P\left( m \right) \text{ is true } . \]
\[\text{ Now } , \]
\[P\left( m + 1 \right) = x^{2m + 1} + y^{2m + 1} \]
\[ = x^{2m + 1} + y^{2m + 1} - x^{2m - 1} . y^2 + x^{2m - 1} . y^2 \]
\[ = x^{2m - 1} \left( x^2 - y^2 \right) + y^2 \left( x^{2m - 1} + y^{2m - 1} \right) \left[ \text{ From } (1) \right]\]
\[ = x^{2m - 1} \left( x^2 - y^2 \right) + y^2 . \lambda\left( x + y \right) \]
\[ = \left( x + y \right)\left( x^{2m - 1} \left( x - y \right) + \lambda y^2 \right) [\text{ It is divisible by } (x + y) . ]\]
\[\text{ Thus, } P\left( m + 1 \right) \text{ is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^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.
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.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
32n+7 is divisible by 8 for all n ∈ N.
(ab)n = anbn 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
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?
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
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:
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/(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:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
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)`
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 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.
A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.
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:
2n < (n + 2)! for all natural number n.
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
