Using principle of mathematical induction, Prove that 1 1 . 2 + 1 2 . 3 + … . . + 1 n ( n + 1 ) = n n + 1 for all natural numbers n .

To prove - 1 1 . 2 + 1 2 . 3 + … . . + 1 n ( n + 1 ) = n n + 1 Step 1 (Verification step) :- For the least natural number i.e n = 1 1 1 1 + 1 = 1 1 + 1 ⇒ 1 2 = 1 2 , which is true. Thus, the result is true for n = 1 . Step 2 (Assumption step) :- For n = k Let the result be true for n = k i.e, 1 1 . 2 + 1 2 . 3 + … . . + 1 k ( k + 1 ) = k k + 1 ⋯ ⋯ 1 Step 3 (Induction step) :- For n = k + 1 To prove - 1 1 . 2 + 1 2 . 3 + … . . + 1 k k + 1 + 1 k + 1 ( k + 2 ) = k + 1 k + 2 We have to simplify L.H.S = 1 1 . 2 + 1 2 . 3 + … . . + 1 k k + 1 + 1 k + 1 ( k + 2 ) = k k + 1 + 1 k + 1 ( k + 2 ) {Using 1 } = k k + 2 + 1 k + 1 ( k + 2 ) = k 2 + 2 k + 1 k + 1 ( k + 2 ) = k 2 + k + k + 1 k + 1 ( k + 2 ) = k k + 1 + 1 k + 1 k + 1 ( k + 2 ) = k + 1 k + 1 k + 1 ( k + 2 ) = k + 1 ( k + 2 ) = R.H.S Thus, the result is true for n = k + 1 . Step 4 (Generalisation step) :- The result is true for n = 1 And the result is true for n = k This implies the result is true for n = k + 1 Therefore, the result is true for all natural numbers.

Using principle of mathematical induction, Prove that 1 . 3 + 2 . 4 + 3 . 5 + … . . + n ( n + 2 ) = n ( n + 1 ) ( 2 n + 7 ) 6 for all natural numbers n .

To prove - 1 . 3 + 2 . 4 + 3 . 5 + … . . + n ( n + 2 ) = n ( n + 1 ) ( 2 n + 7 ) 6 Step 1 (Verification step) :- For the least natural number i.e n = 1 1 1 + 2 = 1 1 + 1 2 1 + 7 6 ⇒ 3 = 2 × 9 6 ⇒ 3 = 3 , which is true. Thus, the result is true for n = 1 . Step 2 (Assumption step) :- For n = k Let the result be true for n = k i.e, 1 . 3 + 2 . 4 + 3 . 5 + … . . + k ( k + 2 ) = k ( k + 1 ) ( 2 k + 7 ) 6 Step 3 (Induction step) :- For n = k + 1 To prove - 1 . 3 + 2 . 4 + 3 . 5 + … . . + k ( k + 2 ) + k + 1 k + 3 = k + 1 ( k + 2 ) ( 2 k + 9 ) 6 We have to simplify L.H.S = 1 . 3 + 2 . 4 + 3 . 5 + … . . + k ( k + 2 ) + k + 1 k + 3 = k k + 1 ( 2 k + 7 ) 6 + k + 1 k + 3 {Using 1 } = k k + 1 ( 2 k + 7 ) + 6 k + 1 k + 3 6 = k + 1 k ( 2 k + 7 ) + 6 k + 3 6 = k + 1 2 k 2 + 13 k + 18 6 = k + 1 2 k 2 + 4 k + 9 k + 18 6 = k + 1 2 k k + 2 + 9 k + 2 6 = k + 1 k + 2 2 k + 9 6 = R.H.S Thus, the result is true for n = k + 1 . Step 4 (Generalisation step) :- The result is true for n = 1 And the result is true for n = k This implies the result is true for n = k + 1 Therefore, the result is true for all natural numbers.

Using principle of mathematical induction, Prove that x n - y n = ( x - y ) x n - 1 + x n - 2 y + … + x y n - 2 + y n - 1 ; x , y ∈ R and n ∈ N . Hint: Write x n + 1 - y n + 1 = x x n - y n + y n ( x - y )

To prove - x n - y n = ( x - y ) x n - 1 + x n - 2 y + … + x y n - 2 + y n - 1 Step 1 (Verification step) :- For the least natural number i.e n = 1 x 1 - y 1 = ( x - y ) ⇒ x - y = x - y , which is true. Thus, the result is true for n = 1 . Step 2 (Assumption step) :- For n = k Let the result be true for n = k i.e, x k - y k = ( x - y ) x k - 1 + x k - 2 y + … + x y k - 2 + y k - 1 ⋯ ⋯ 1 Step 3 (Induction step) :- For n = k + 1 To prove - x k + 1 - y k + 1 = ( x - y ) x k + x k - 1 y + … + x y k - 1 + y k We have to simplify L.H.S = x k + 1 - y k + 1 = x k x - y k y = x k x - x y k + x y k - y k y = x x k - y k + y k x - y = x ( x - y ) x k - 1 + x k - 2 y + … + x y k - 2 + y k - 1 + y k x - y {Using 1 } = x - y x x k - 1 + x k - 2 y + … + x y k - 2 + y k - 1 + y k = x - y x k + x k - 1 y + … + x y k - 1 + y k = R.H.S Thus, the result is true for n = k + 1 . Step 4 (Generalisation step) :- The result is true for n = 1 And the result is true for n = k This implies the result is true for n = k + 1 Therefore, the result is true for all natural numbers.

Using principle of mathematical induction, Prove that 1 + 3 + 5 + … + ( 2 n - 1 ) = n 2 for all natural numbers n .

To prove - 1 + 3 + 5 + … . . + ( 2 n - 1 ) = n 2 Step 1 (Verification step) :- For the least natural number i.e n = 1 2 1 - 1 = 1 2 ⇒ 1 = 1 , which is true. Thus, the result is true for n = 1 . Step 2 (Assumption step) :- For n = k Let the result be true for n = k i.e, 1 + 3 + 5 + … . . + ( 2 k - 1 ) = k 2 Step 3 (Induction step) :- For n = k + 1 To prove - 1 + 3 + 5 + … . . + ( 2 k - 1 ) + 2 k + 1 = k + 1 2 We have to simplify L.H.S = 1 + 3 + 5 + … . . + ( 2 k - 1 ) + 2 k + 1 = k 2 + 2 k + 1 {Using 1 } = k 2 + k + k + 1 = k k + 1 + 1 k + 1 = k + 1 k + 1 = k + 1 2 = R.H.S Thus, the result is true for n = k + 1 . Step 4 (Generalisation step) :- The result is true for n = 1 And the result is true for n = k This implies the result is true for n = k + 1 Therefore, the result is true for all natural numbers.

Prove the following by mathematical induction: ( 1 · 2 · 3 … , n ) 3 > 8 1 3 + 2 3 + 3 3 + … + n 3 , for n > 3

Let P ( n ) ≡ ( 1 · 2 · 3 ⋯ n ) 3 > 8 1 3 + 2 3 + 3 3 + ⋯ + n 3 , for n > 3 P ( 4 ) ≡ ( 1 · 2 · 3 · 4 ) 3 > 8 1 3 + 2 3 + 3 3 + 4 3 ⇒ ( 24 ) 3 > 8 ( 1 + 8 + 27 + 64 ) ⇒ 13824 > 8001 P ( 4 ) is true. Let P ( k ) is true. ⇒ ( 1 · 2 · 3 · ⋯ · k ) 3 > 8 1 3 + 2 3 + 3 3 + ⋯ + k 3 For n = k + 1 , P ( k + 1 ) ≡ ( 1 · 2 · 3 ⋯ k ( k + 1 ) ) 3 > 8 1 3 + 2 3 + 3 3 + ⋯ + k 3 + ( k + 1 ) 3 ⇒ ( 1 . 2 · 3 ⋯ k ) 3 ( k + 1 ) 3 > 8 1 3 + 2 3 + 3 3 + ⋯ + k 3 + 8 ( k + 1 ) 3 ∴ P ( k + 1 ) is true Because if a > b , then a c > b + c where a , b , c > 1 P ( 4 ) is true and P ( k + 1 ) is true whenever P ( k ) is true. ∴ By the principle of mathematical induction, P ( n ) is true for n > 3 .

Prove the following by induction: 1 n + 1 + 1 n + 2 + … + 1 3 n + 1 > 1 for every positive integer n .

Let P n ≡ 1 n + 1 + 1 n + 2 + ⋯ + 1 3 n + 1 > 1 P 1 ≡ 1 2 + 1 3 + 1 4 > 1 ⇒ 13 12 > 1 ∴ P 1 is true. Let P ( k ) is true. ⇒ 1 k + 1 + 1 k + 2 + ⋯ + 1 3 k + 1 > 1 For n = k + 1 , LHS = 1 k + 1 + 1 k + 2 + ⋯ + 1 3 k + 1 + 1 3 ( k + 1 ) + 1 > 1 + 1 3 k + 5 > 1 ∴ P ( k + 1 ) is true P ( 1 ) is true and P ( k + 1 ) is true whenever P ( k ) is true. ∴ By the principle of mathematical induction, P ( n ) is true for every positive integer n .

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 5 - Principle of Mathematical Induction>Exercise-5

Odisha Board Solutions for Chapter: Principle of Mathematical Induction, Exercise 1: Exercise-5

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Principle of Mathematical Induction, Exercise 1: Exercise-5

Attempt the free practice questions on Chapter 5: Principle of Mathematical Induction, Exercise 1: Exercise-5 with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.

Questions from Odisha Board Solutions for Chapter: Principle of Mathematical Induction, Exercise 1: Exercise-5 with Hints & Solutions

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 5 - Principle of Mathematical Induction>Exercise-5>Q 9

Using principle of mathematical induction, Prove that $5^{2 n + 2} - 24 n - 25$ is divisible by 576 for all natural numbers $n .$

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 5 - Principle of Mathematical Induction>Exercise-5>Q 10

Using principle of mathematical induction, Prove that $\frac{1}{1.2} + \frac{1}{2.3} + \dots . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}$ for all natural numbers $n$ .

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 5 - Principle of Mathematical Induction>Exercise-5>Q 11

Using principle of mathematical induction, Prove that $1.3 + 2.4 + 3.5 + \dots . . + n (n + 2) = \frac{n (n + 1) (2 n + 7)}{6}$ for all natural numbers $n$ .

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 5 - Principle of Mathematical Induction>Exercise-5>Q 12

Using principle of mathematical induction, Prove that $x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} + y^{n - 1}); x, y \in R$ and $n \in N$ .
$[$ Hint: Write $x^{n + 1} - y^{n + 1} = x (x^{n} - y^{n}) + y^{n} (x - y)]$