English

1.2 + 2.3 + 3.4 + ... + N (N + 1) = N ( N + 1 ) ( N + 2 ) 3 - Mathematics

Advertisements
Advertisements

Question

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

 
Advertisements

Solution

Let P(n) be the given statement.
Now,

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

\[\text{ Step }  1: \]

\[P(1) = 1 . 2 = 2 = \frac{1(1 + 1)(1 + 2)}{3}\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step } 2: \]

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

\[\text{ Then, } \]

\[1 . 2 + 2 . 3 + . . . + m(m + 1) = \frac{m(m + 1)(m + 2)}{3}\]

\[\text{ To prove: P(m + 1) is true .}  \]

\[\text{ That is,}  \]

\[1 . 2 + 2 . 3 + . . . + (m + 1)(m + 2) = \frac{(m + 1)(m + 2)(m + 3)}{3}\]

\[\text{ Now, P(m) is }  \]

\[1 . 2 + 2 . 3 + . . . + m(m + 1) = \frac{m(m + 1)(m + 2)}{3}\]

\[ \Rightarrow 1 . 2 + 2 . 3 + . . . + m(m + 1) + (m + 1)(m + 2) = \frac{m(m + 1)(m + 2)}{3} + (m + 1)(m + 2)\]

\[ \Rightarrow P(m + 1) = \frac{(m + 1)(m + 2)(m + 3)}{3}\]

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 14 | Page 27

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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+ 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

`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

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

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.


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


If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

 

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

 

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

 


\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\] 


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

 

1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]

 

a + ar + ar2 + ... + arn−1 =  \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]

 

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

 


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


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

 


\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{  is true for all n }  \in N?\]


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


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

 


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

2 + 4 + 6 + ..... + 2n = n (n+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:

(24n−1) is divisible by 15


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

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)


Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.


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. Justify your answer. 


Prove the statement by using the Principle of Mathematical Induction:

23n – 1 is divisible by 7, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, 7n – 2n is divisible by 5.


Prove the statement by using the Principle of Mathematical Induction:

n(n2 + 5) is divisible by 6, for each natural number n.


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 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers 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 for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin  beta/2)`.


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


Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×