Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): \frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n \text{ is a positive integer for all } n \in N . \]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{62}{165} = \frac{15 + 33 + 55 + 62}{165} = \frac{165}{165} = 1 \]
\[\text{ It is certainly a positive integer } . \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then, } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \text{ is a positive integer . } \]
\[\text{ Now, let } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m = \lambda, \text{ where } \lambda \in N\text{ is a positive integer . } \]
\[\text{ We have to show that P(m + 1) is true whenever P(m) is true } . \]
\[\text{ To prove: } \frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\text{ is a positive integer .} \]
\[\text{ Now, } \]
\[\frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\]
\[ = \frac{1}{11}\left( m^{11} + 11 m^{10} + 55 m^9 + 165 m^8 + 330 m^7 + 462 m^6 + 462 m^5 + 330 m^4 + 165 m^3 + 55 m^2 + 11m + 1 \right)\]
\[ + \frac{1}{5}\left( m^5 + 5 m^4 + 10 m^3 + 10 m^2 + 5m + 1 \right) + \frac{1}{3}\left( m^3 + 3 m^2 + 3m + 1 \right)\]
\[ + \frac{62}{165}m + \frac{62}{165}\]
\[ = \left[ \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \right] + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{6}{105}\]
\[ = \lambda + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + 1\]
\[\text{ It is a positive integer } . \]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
`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: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^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: 32n + 2 – 8n– 9 is divisible by 8.
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 "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
n(n + 1) (n + 5) is a multiple of 3 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?
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
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:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
Prove by method of induction, for all n ∈ N:
`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`
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, for all n ∈ N
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
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
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.
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
Prove the statement by using the Principle of Mathematical Induction:
For any natural number n, 7n – 2n is divisible by 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 ______.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
