मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove by method of induction, for all n ∈ N: 3 + 7 + 11 + ..... + to n terms = n(2n+1) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

3 + 7 + 11 + ..... + to n terms = n(2n+1)

बेरीज
Advertisements

उत्तर

Let P(n) ≡ 3 + 7 + 11 + ... to n terms = n(2n + 1), for all n ∈ N.

3, 7, 11, ... are in A.P. with a = 3, d = 4

∴ nth term = a+ (n – 1)d = 3 + (n – 1)4 = 4n – 1

∴ P(n) ≡ 3 + 7 + 11 + ... + (4n – 1) = n (2n + 1)

Step 1: For n = 1, L.H.S. = 3

R.H.S. = 1(2 × 1 + 1) = 3

∴ L.H.S. = R.H.S. for n = 1

∴ P(1) is true.

Step 2: Let us assume that for some k ∈ N, P(k) is true, i.e., 3 + 7 + 11 + ... + (4k – 1) = k(2k + 1)  ...(1)

Step 3: To prove that P(k + 1) is true, i.e., to prove that 3 + 7 + 11 + ... + (4k – 1) + (4k + 3) = (k + 1)(2k + 3)

Now, L.H.S. = 3 + 7 + 11 + ... + (4k – 1) + (4k + 3)

= k(2k + 1) + (4k + 3) ... [By (1)]

= 2k2 + k + 4k + 3

= 2k2 + 3k + 2k + 3

= k(2k + 3) + 1(2k + 3)

= (k + 1)(2k + 3)

= R.H.S

∴ P(k + 1) is true.

Step 4: From all the above steps and by the principle of mathematical induction, the result P(n) is true for all n ∈ N, i.e., 3 + 7 + 11 + ... to n terms = n(2n + 1), for all n ∈ N.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [पृष्ठ ७३]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 2 | पृष्ठ ७३

संबंधित प्रश्‍न

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+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: `1+2+ 3+...+n<1/8(2n +1)^2`


Prove the following by using the principle of mathematical induction for all n ∈ Nx2n – y2n is divisible by x y.


Prove the following by using the principle of mathematical induction for all n ∈ N (2+7) < (n + 3)2


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

 

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


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

 

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

 

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


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

 

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


\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\]  for all n ∈ N .


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

 


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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

 


\[\text{ Using principle of mathematical induction, prove that } \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]

 


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:

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


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:

(23n − 1) is divisible by 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:

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:

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


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 ______.


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:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.


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 number of subsets of a set containing n distinct elements is 2n, 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 ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×