Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.

1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)

Solution:

Let p_{n} be the given statement

Question 2.

1^{2} + 2^{2} + …… + n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\)

Solution:

Let p_{n} be the given statement

when n = 1

1^{2} =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)

P_{1} is true

Let P_{k} be true.

i.e. 1^{2} + 2^{2} + … + k^{2} = \(\frac{k(k+1)(2 k+1)}{6}\)

we shall prove P_{k} + 1 is true i.e., 1^{2} + 2^{2} + … + k^{2} + (k + 1)^{2}

\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)

∴ P_{k+1} is true

∴ P_{n} is true of all values of n∈N.

Question 3.

1 + r + r^{2}+ …. + r^{n} = \(\frac{r^{n+1}-1}{r-1}\)

Solution:

Let P_{n} be the given statement

Question 4.

5^{n} – 1 is divisible by 4.

Solution:

Let P_{n} = 5^{n }– 1

When n = 1,

5^{1} – 1 = 4 is divisible by 4.

∴ P_{1}, is true.

Let P_{k} be true i.e.,

5^{k }– 1 is divisible by 4.

Let 5^{k+1 }– 1 = 4m, me Z

Now 5k + 1 – 1 = 5^{k}. 5 – 5 + 4

= 5 (5^{k} – 1) + 4

= 5 × 4m + 4 = 4 (5m + 1)

which is divisible by 4.

∴ P_{k+1} is true.

∴ P_{n} is true for all values of n ∈ N

Question 5.

7^{2n} + 2^{3n-3} 3^{n-1} is divisible by 25 for any natural number n > 1.

Solution:

Let 7^{2n} + 2^{3n-3} . 3^{n-1}

when n = 1, 7^{1} + 2^{0} . 3^{0} ⇒ 49 + 1 = 50

which is divisible by 25.

∴ P_{1} is true. Let P_{k} be true.

i.e., 72^{k }+ 2^{3k-3} 3^{k-1} is divisible by 35

Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)

=7^{2k+2} + 2^{3k}. 3^{k}

= 7^{2k}. 7^{2} + 2^{3k-3} 2^{3}. 3^{k-1}. 3^{1}

= 7^{2k}. 49 + 2^{3k-3}. 3^{k-1}. 24

= 7^{2k} (25 + 24) + 24. 2^{3k-3}. 3^{k-1}

= 7^{2k}. 25 + 24 (7^{2k} + 2^{3k-3}. 3^{k-1})

= 7^{2k}. 25 + 24 × 25m

Which is divisible by 25 (∵ P_{k} is true)

∴ P_{k+1} is true

∴ P_{n }is true for all values of n > 1.

Question 6.

7. 5^{2n-1} + 23^{n+1} is divisible by 17 for every natural number n ≥ 1.

Solution:

Let P_{n} = 7. 5^{2n} – 1 + 2^{3n+1}.

When n = 1, 7.5 + 2^{4} = 35 + 16 = 51

Which is divisible by 17.

P_{1}, is true.

Let P_{k} be true i.e., 7.5^{2k-1} + 2^{3k+1 } is divisible by 17.

Let 7.5^{2k-1} + 2^{3k+1} = 17 m, m ∈ Z

Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)

= 7.5^{2k-1} + 2^{3k+4 }

= 7.5^{2k-1} . 5^{2} + 2^{3k+1} . 2^{3}

= 25. 7. 5^{2k-1} + 8. 2^{3k+1}

= (17 + 8) 7.5^{2k-1} + 8. 2^{3k+1}

= 17. 7. 5^{2k-1} + 8 (7. 5^{2k-1} + 2^{3k+1})

= 17 × 7 × 5^{2k-1} + 8 × 17m

Which is divisible by 17.

Hence P_{k+1}. is true.

∴ P_{n} is true for all values of n ≥ 1.

Question 7.

4^{n+1} + 15n + 14 is divisible by 9 for every natural number n ≥ 0.

Solution:

Let P_{n} = 4^{n+1}+ 15 n + 14

when n = 1, 4^{2} + 15 + 14 = 45 is divisible by 9.

∴ P_{1} is true. Let P_{k} be true.

i.e., 4^{k+1 }+ 15k + 14 is divisible by 9.

Now, 4^{k+1+1 }+ 15 (k + 1) + 14

= 4^{k+2} + 15k + 29

= 4^{k+1}. 4 + 60k + 56 – 45k – 27

= 4 (4^{k+1} + 15k + 14) – 9 (5k + 3)

Which is divisible by 9.

∴ P_{k+1}, is true.

∴ P_{n} is true for all values of n ≥ 0.

Question 8.

3^{(2n-1)} + 7 is divisible by 9 for every natural number n ≥ 2.

Solution:

Let P_{n }= 3^{2(n-1)} + 7

When n = 2. 3^{2} + 7 = 16 is divisible by 8.

∴ P_{2} is true.

Let P_{k} be true.

i.e., 5^{2(k-1)} + 7 is divisible by 8.

Let 3^{2k-2} + 7 = 8m. m ∈ Z.

Now \(3^{2(\overline{k+1}-1)}\) + 7 = 3^{2k} + 7

= 3^{2k-2}. 3^{2} + 63 – 56

= 9(3^{2k-2} + 7) – 56

= 9 × 8m – 56 = 8 (9m – 7)

Which is divisible by 8.

P_{k+1} is true.

P_{n} is true for all values on n ≥ 2.

Question 9.

5^{(2n-4) }– 6n + 32 is divisible by 9 for every natural number n ≥ 5.

Solution:

Let P = 5^{2(n-4)} – 6n + 32

For n = 5, P_{5} = 5^{2} – 6. 5 + 32

= 25 – 30 + 32 = 27

Which is divisible by 9.

Hence P_{5} is true.

Let P_{k} is true.

Let P_{k} is divisible by 9.

Let P_{k} = 5^{2(k-4)} – 6k + 32 = 9., m ≥ Z

5^{2k+2} = 576 m + 24k + 25 … (1)

we shall prove that P_{k+1} is true.

Now 5^{2(K+1)+2} – 24(k+1) – 25

= 5^{2} (5^{2k+2}) – 24k – 24 – 25

= 5^{2}[576m + 24k + 25] – 24k – 24 – 25

= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25

= 25 × 576 m + 576 k + 576

= 576 [25 m + k + 1]

which is divisible by 576

∴ P_{k+1} is true.

So by the method of induction P_{n} is true for all n.

i.e., 5^{2n+2} – 24n – 25 is divisible by 576 for all n ∈ N.

Hence P_{k+1} is true.

So by methods of induction P_{n} is true.

i.e., 5^{2n+2} – 24n – 25 is divisible by 576 for all n.

Question 10.

\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)

Solution:

when n = 1,

Question 11.

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

Solution:

when n = 1,

we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)

\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)

Question 12.

x^{n} – y^{n} = (x – y)(x^{n-1} + x^{n-2} y + … + xy^{n-2} + y^{n-1}); x, y ∈ R [Hint : Write x^{n+1} – y^{n+1} = x(x^{n} – y^{n}) + y^{n}(x – y)]

Solution:

Let p(n) is

x^{n} – y^{n} = (x – y)(x^{n-1} + x^{n-2} y + … + xy^{n-2} + y^{n-1}); x, y ∈ R

Step – 1:

For n = 2

x^{2} – y^{2} = (x – y) (x + y) (True)

∴ P(1) is true.

Step – 2:

Let P(k) is true.

i.e., x^{k} – y^{k} = (x – y)(x^{k-1} + x^{k-2}y + … +xy^{k-2} + y^{k-1})

Step – 3:

Let us prove P_{k+1} is true.

i.e., x^{k+1} – y^{k+1} = (x – y) (x^{k} + x^{k-1}y + … (xy^{k-1} + y^{k})

L.H.S. = x^{k+1} – y^{k+1}

= x^{k+1} – xy^{k} + xy^{k} – y^{k+1}

= x(x^{k }– y^{k}) + y^{k} (x – y)

= x(x – y)(x^{k-1} + x^{k-2} y + … + xy^{k-2} + y^{k-1}) + y^{k}(x – y) [by (1)]

= (x – y) [x^{k} + x^{k-1} y + … + xy^{k-2} + xy^{k-1} + y^{k}]

= R.H.S.

∴ P_{(k+1)} is true.

Step – 4:

By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.

1 + 3 + 5 + ……. +(2n – 1) = n^{2}

Solution:

Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n^{2}.

Step – 1:

For n = 2

L.H.S. = 1 + 3 = 4 = 2^{2} (R.H.S)

∴ P(1) is true.

Step – 2:

Let P(k) is true.

i.e., 1 + 3 + 5 … + (2k – 1) = k^{2} …(1)

Step – 3:

We will prove that P(k + 1) is true

i.e., we want to prove.

1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)^{2}

L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k^{2} + 2k + 1 [By – (1)]

= (k + 1)^{2} = R.H.S.

Step – 4:

By the Principle of Mathematical Induction P(n) is true for all n.

i.e., 1 + 3 + 5 ….+ (2n – 1) = n^{2}

Question 14.

n > n; n is a natural number.

Solution:

Let P(n) is 2^{n} > n

Step – 1:

2^{1} > 1 (True)

∴ P(1) is true.

Step – 2:

Let P(k) is true.

⇒ 2^{k} > k

Step – 3:

We shall prove that P(k + 1) is true

i.e., 2^{k+1} > k + 1

Now 2^{k+1} = 2.2k > 2k ≥ k + 1 for k ∈ N.

∴ 2^{k+1} > k + 1

⇒ P(k + 1) is true.

Step – 4:

By the Principle of Mathematical Induction P(n) is true for all n.

i.e., 2^{n} > n for n ∈ N

Question 15.

(1, 2, 3 … n)^{3} > 8 (1^{3} + 2^{3} + 3^{3} + … + n^{3}), for n > 3.

Solution:

Let P(n) is

(1, 2, 3 … n)^{3} > 8 (1^{3} + 2^{3} + 3^{3} + … + n^{3}), for n > 3.

Step – 1:

For n = 4

(1. 2. 3. 4)^{3} = 24^{3} = 13824

8(1^{3}+ 2^{3} + 3^{3} + 4^{3}) = 808

∴ (1. 2 . 3 . 4)^{3} > 8(1^{3} + 2^{3} + 3^{3} + 4^{3})

∴ P(4) is true.

Step- 2:

Let P(k) is true.

(1. 2. 3…….k)^{3} > 8(1^{3} + 2^{3} + 3^{3} + …… + k^{3})

Step – 3:

We shall prove that P_{(k+1)} is true.

i.e., (1. 2. 3. …….. k(k+1))^{3} > 8(1^{3} + 2^{3} + … + k^{3} + (k + 1)^{3})

Now (1. 2. 3. …….. k(k + 1)^{3})

= (1. 2. 3 … k)^{3} (k + 1)^{3}

> 8 (1^{3} + 2^{3} + … k^{3}) (k + 1)^{3}

> 8 (1^{3} + 2^{3} + … k^{3}) + 8(k + 1)^{3}

= 8 (1^{3} + 2^{3} + … + k^{3} + (k + 1)^{3})

P_{(k+1)} is true.

Step – 4:

By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.

\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.

Solution:

Let P(n) is

\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)

Step-4:

By the Principle of Mathematical Induction P(n) is true for all n ∈ N.