हिंदी

N(N + 1) (N + 5) is a Multiple of 3 for All N ∈ N. - Mathematics

Advertisements
Advertisements

प्रश्न

n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

 

उत्तर

Let P(n) be the given statement.
Now,

\[P(n): n(n + 1)(n + 5)\text{ is a multiple of }  3 . \]

\[\text{ Step1 } : \]

\[P(1): 1(1 + 1)(1 + 5) = 12 \]

\[\text{ It is a multiple of}  3 . \]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step2 } : \]

\[\text{ Let } P\left( m \right) \text{ be true .  } \]

\[\text{ Then,}  m\left( m + 1 \right)\left( m + 5 \right) \text{ is a multiple of } 3 . \]

\[\text{ Suppose}  m\left( m + 1 \right)\left( m + 5 \right) = 3\lambda, \text{ where }  \lambda \in N . \]

\[\text{ We have to show that } P\left( m + 1 \right) \text{ is true whenever P(m) is true } . \]

\[\text{ Now, } \]

\[P(m + 1) = \left( m + 1 \right)\left( m + 2 \right)\left( m + 6 \right)\]

\[ = m\left( m + 1 \right)\left( m + 6 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]

\[ = m\left( m + 1 \right)\left( m + 5 + 1 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]

\[ = m\left( m + 1 \right)\left( m + 5 \right) + m(m + 1) + 2\left( m + 1 \right)\left( m + 6 \right)\]

\[ = 3\lambda + \left( m + 1 \right)\left( m + 2m + 6 \right) \left[ From P(m) \right]\]

\[ = 3\lambda + 3\left( m + 1 \right)\left( m + 2 \right)\]

\[\text{ It is clearly a multiple of 3}  . \]

\[\text{ Thus, P(m + 1) is true } . \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n }   \in N .\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 24 | पृष्ठ २८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Prove the following by using the principle of mathematical induction for all n ∈ N

`1+ 1/((1+2)) + 1/((1+2+3)) +...+ 1/((1+2+3+...n)) = (2n)/(n +1)`

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+ 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: 41n – 14n is a multiple of 27.


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 + 2 + 3 + ... + n =  \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .


1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]

 

1.2 + 2.22 + 3.23 + ... + n.2= (n − 1) 2n+1+2

 

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

 

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]


32n+7 is divisible by 8 for all n ∈ N.

 

2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.


Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all \[\in\] N .

 

7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]

 

\[\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n\] is a positive integer for all n ∈ N

 


\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ and  \[0 < x < \frac{\pi}{2}\]

 


\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\] for all n ≥ 2, n ∈ 

 


x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.

 

\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 


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:

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)


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/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`


Prove by method of induction, for all n ∈ N:

(23n − 1) is divisible by 7


Prove by method of induction, for all n ∈ N:

(24n−1) is divisible by 15


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))`


Answer the following:

Prove by method of induction

`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀  "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:

`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2.


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))`


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:

n3 – 7n + 3 is divisible by 3, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

n3 – n is divisible by 6, for each natural number n ≥ 2.


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, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin  ((n + 1))/2 theta)/(sin  theta/2)`, for all n ∈ N.


For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.


If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.


By using principle of mathematical induction for every natural number, (ab)n = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×