English

1.3 + 2.4 + 3.5 + ... + N. (N + 2) = 1 6 N ( N + 1 ) ( 2 N + 7 )

Advertisements
Advertisements

Question

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

 
Advertisements

Solution

Let P(n) be the given statement.
Now, 

\[P(n) = 1 . 3 + 2 . 4 + 3 . 5 + . . . + n . (n + 2) = \frac{1}{6}n(n + 1)(2n + 7)\]

\[\text{ Step } 1: \]

\[P(1) = 1 . 3 = 3 = \frac{1}{6} \times 1(1 + 1)(2 \times 1 + 7)\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2:}  \]

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

\[\text{ Then, } \]

\[1 . 3 + 2 . 4 + . . . + m . (m + 2) = \frac{1}{6}m(m + 1)(2m + 7)\]

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

\[\text{ That is, } \]

\[1 . 3 + 2 . 4 + . . . + (m + 1)(m + 3) = \frac{1}{6}(m + 1)(m + 2)(2m + 9)\]

\[P(m) \text{ is equal to }  1 . 3 + 2 . 4 + . . . + m(m + 2) = \frac{1}{6}m(m + 1)(2m + 7) . \]

\[\text{ Thus, we have: }  \]

\[1 . 3 + 2 . 4 + . . . + m(m + 2) + (m + 1)(m + 3) = \frac{1}{6}m(m + 1)(2m + 7) + (m + 1)(m + 3) \left[ \text{ Adding } (m + 1)(m + 3)\text{  to both sides } \right]\]

\[ \Rightarrow 1 . 3 + 2 . 4 + . . . + (m + 1)(m + 3) = \frac{1}{6}(m + 1)\left[ 2 m^2 + 7m + 6m + 18 \right]\]

\[ = \frac{1}{6}(m + 1)(2 m^2 + 13m + 18)\]

\[ = \frac{1}{6}(m + 1)(2m + 9)(m + 2)\]

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

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

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

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 12 | Page 27

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: 41n – 14n is a multiple of 27.


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

 

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

 

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


1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

 

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


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

 

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

 

32n+2 −8n − 9 is divisible by 8 for all n ∈ N.


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.

 

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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]

 

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

 

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

 


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


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

 

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 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`


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:

(24n−1) is divisible by 15


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

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1


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

`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N


Answer the following:

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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`


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, for all n ∈ N

2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n 


Answer the following:

Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2


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 statement by using the Principle of Mathematical Induction for all n ∈ N, that:

2n + 1 < 2n, for all natual numbers n ≥ 3.


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


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


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:

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 that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×