English

Prove the Following by Using the Principle of Mathematical Induction for All N ∈ N: 1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = (N(4n^2 + 6n -1))/3

Advertisements
Advertisements

Question

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

1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = `(n(4n^2 + 6n -1))/3`
Advertisements

Solution

Let the given statement be P(n), i.e.,

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

shaalaa.com
  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

(1+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`

 

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

`1^2 + 3^2 + 5^2 + ... + (2n -1)^2 = (n(2n - 1) (2n + 1))/3`

Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11


Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.


Given an example of a statement P (n) such that it is true for all n ∈ N.

 

If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.


12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .

 

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

 


\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]


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

 

2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]

 

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


12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]

 

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


11n+2 + 122n+1 is divisible by 133 for all n ∈ N.

 

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

 

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

 


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


\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

 


Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .

 

\[\text{ A sequence }  a_1 , a_2 , a_3 , . . . \text{ is defined by letting }  a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]


\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{  for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n }  \in N \text{ using mathematical induction .} \]


\[\text{ The distributive law from algebra states that for all real numbers}  c, a_1 \text{ and }  a_2 , \text{ we have }  c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]


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:

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`


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:

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 152n–1 + 1 is divisible by 16, for all 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.


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"):}`


Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.


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.


Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.


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


Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×