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: 1.2.3 + 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 + 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:
`(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:
Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
52n+2 −24n −25 is divisible by 576 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 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
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?
x2n−1 + y2n−1 is divisible by x + y for all n ∈ 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:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 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:
Prove by method of induction loga xn = n logax, x > 0, n ∈ N
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
22n – 1 is divisible by 3.
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.
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:
n2 < 2n for all natural numbers n ≥ 5.
Prove the statement by using the Principle of Mathematical Induction:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
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.
A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, 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 ______.
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.
