# CBSE Solutions for Class 11 English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

State whether the following is true or false :  12 + 32 + 52 + ... + (2n − 1)2 = n(4n2−1)/3

જવાબ : True

State whether the following is true or false :  ½ + ¼ + 1/8 +...+ 1/2n = 1 – 1/2n

જવાબ : True

State whether the following is true or false :  1.2 + 2.3 + 3.4 + ... + n (n + 1) = n(n+1)(n+2)/3

જવાબ : True

State whether the following is true or false :  1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) = n(4n2+6n−1)/3

જવાબ : True

State whether the following is true or false :  1.3 + 2.4 + 3.5 + ... + n. (n + 2) = n(n+1)(2n+7)/6

જવાબ : True

State whether the following is true or false :  2 + 5 + 8 + 11 + ... + (3n − 1) = ½ n(3n+1)

જવાબ : True

State whether the following is true or false :  1.2 + 2.22 + 3.23 + ... + n.2= (n − 1) 2n+1+2

જવાબ : True

State whether the following is true or false :  52n −1 is divisible by 24 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  32n+7 is divisible by 8 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  52n+2 −24n −25 is divisible by 576 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  32n+2 −8n − 9 is divisible by 8 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  (ab)n = anbn for all n ∈ N.

જવાબ : True

State whether the following is true or false :  n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  11n+2 + 122n+1 is divisible by 133 for all n ∈ N.

જવાબ : True

State whether the following is true or false :  12 + 32 + 52 + ... + (2n − 1)2 =  n(4n2−1)/3

જવાબ : True

State whether the following is true or false :  Prove that n3 −- 7+ 3 is divisible by 3 for all n ∈∈ N.

જવાબ : True

State whether the following is true or false :  Prove that 1 + 2 + 22 + ... + 2n = 2n+1 − 1 for all n ∈∈ N.

જવાબ : True

State whether the following is true or false :  7 + 77 + 777 + ... + 777  ...........n−digits7 = 7(10n+1−9n−10)/81

જવાબ : True

State whether the following is true or false :  1/3.7 + 1/7.11 + 1/11.5 +...+ 1/(4n−1)(4n+3) = n/3(4n+3)

જવાબ : True

State whether the following is true or false :  1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(2n+1)(2n+3) = n/3(2n+3)

જવાબ : True

State whether the following is true or false :  1/1.4 + 1/4.7 + 1/7.10 +...+ 1/(3n−2)(3n+1) = n/3n+1

જવાબ : True

State whether the following is true or false :  n7/7+n5/5+n3/3+n2/2−37n/210 is a positive integer for all n ∈ N.

જવાબ : True

State whether the following is true or false :  n11/11+n5/5+n3/3+62n/165 is a positive integer for all n ∈ N.

જવાબ : True

State whether the following is true or false :  ½ tan(x/2) + ¼ tan(x/4) + ... + (½)n tan(x/2n) = (½)n cot(x/n2) − cot x for all n ∈ and 0

જવાબ : True

State whether the following is true or false :  If 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?

જવાબ : False

State whether the following is true or false : ≤ (1/3n+1)1/2 for all n ∈ N

જવાબ : True

State whether the following is true or false :  1+1/4+1/9+1/16+...+1/n2 < 2−1/n for all n ≥ 2, n ∈ N

જવાબ : True

State whether the following is true or false :  x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.

જવાબ : True

State whether the following is true or false :  1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

જવાબ : True

State whether the following is true or false :  12.5+15.8+18.11+...+1(3n−1)(3n+2) = n/(6n+4)

જવાબ : True

If P(n) : n2 – n is divisible by 2, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 1

If P(n) : n3 – n is divisible by 6, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 1

If P (n) is the statement "n2 − n + 41 is prime", prove that false for P (41)

જવાબ : P(41)=412−41+41=1681 (not prime)

If P (n) is the statement "n2 − n + 41 is prime", prove that P (3)

જવાબ : P(3) =32−3+41=9−3+41=47 (prime)

If P (n) is the statement "n2 − n + 41 is prime", prove that P (2)

જવાબ : P(2)=22−2+41 =4−2+41 =43  (prime)

If P (n) is the statement "n2 − n + 41 is prime", prove that P (1)

જવાબ : P(1) =12−1+41=41  (prime)

If P(n) : 7n < n7n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 8

If P(n) : 6n < n6n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 7

If P(n) : 5n < n5n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 6

If P(n) : 4n < n4n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 5

If P(n) : 3n < n3 , n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 4

If P(n) : 6n < n!, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 4

If P(n) : 5n < n!, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 3

If P(n) : 4n < n!, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 3

If P(n) : 3n < n!, n ∈ N, then P(n) is true for all n ≥ _____________.

જવાબ : 3

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

જવાબ : 4

Prove that -6n < n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -6n < n

If n =1, -61>1. Hence P(1) is true

If n =2, -62 = 36 > 2. Hence P(2) isn’t true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that -11n < n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -11n < n

If n =1, -111>1. Hence P(1) is true

If n =2, -112 = 121>2 . Hence P(2) isn’t true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that -√2n < n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -√2n < n

If n =1, -√21>1. Hence P(1) is true

If n =2, -√22 = 2. Hence P(2) isn’t true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that -2n < n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -2n < n

If n =1, -21>1. Hence P(1) is true

If n =2, -22>1 = 4 > 1. Hence P(2) isn’t true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that -23n < n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -23n < n

If n =1, -231>1. Hence P(1) is true

If n =2, -232>1 = 529 > 1. Hence P(2) isn’t true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that 23n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 23n > n

If n =1, 231>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

23k > k …(1)

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

Now, multiplying both sides of the equation (1) by 23, we get

23 x 23k > 23k

Now by using the property,

i.e., 23k+1> 23k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that 13n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 13n > n

If n =1, 131>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

13k > k …(1)

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

Now, multiplying both sides of the equation (1) by 13, we get

13 x 13k > 13k

Now by using the property,

i.e., 13k+1> 13k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that 2n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 2n > n

If n =1, 2>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

2k > k …(1)

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

Now, addinging both sides of the equation (1) by 1, we get

2k + 1> k+1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that √3n > 2n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): √3n > 2n

If n =1, √31>2. Hence P(1) isn’t  true

Therefore, P(n) isn’t  true for every positive integer n is proved using the principle of mathematical induction.

Prove that 3n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 3n > n

If n =1, 31>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

3k > k …(1)

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

Now, multiplying both sides of the equation (1) by 3, we get

3 x 3k > 3k

Now by using the property,

i.e., 3k+1> 3k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that 1 + 3 + 5 + … + (2n – 1) = n2 using the principle of Mathematical induction.

જવાબ : Given, 1 + 3 + 5 + … + (2n – 1) = n2

Let, P(n) : 1 + 3 + 5 +…+ (2n – 1) = n2 , for n ∈ N

P(1) is true, as

P(1) : 1 = 12

Let as assume P(k) is true for some k ∈ N,

It means that,

P(k) : 1 + 3 + 5 + … + (2k – 1) = k2

To prove that P(k + 1) is true, we have

1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k2 + (2k + 1)

= k2 + 2k + 1

= (k + 1)2

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

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

જવાબ : P(n) be the given statement, that is

P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n.

P (1) is true, as

P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1.

Let P(n) is true for natural number k,

Hence,

P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1

to prove P (k + 1) is true, we have

P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)!

= (k + 1)! – 1 + (k + 1)! × (k + 1)

= (k + 1 + 1) (k + 1)! – 1 = (k + 2) (k + 1)! – 1 = ((k + 2)! – 1

Therefore, P (k + 1) is true, whenever P (k) is true.

Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Induction

Prove that 5n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 5n > n

If n =1, 51>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

5k > k …(1)

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

Now, multiplying both sides of the equation (1) by 5, we get

5 x 5k > 5k

Now by using the property,

i.e., 5k+1 > 5k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that 2 + 4 + 6 + … + (2n) = n(n+1) using the principle of Mathematical induction.

જવાબ : Given, 2 + 4 + 6 + … + (2n) = n(n+1)

Let, P(n) : 2 + 4 + 6 + … + (2n) = n(n+1) , for n ∈ N

P(1) is true, as

P(1) : 2 = 1(1+1)

Let as assume P(k) is true for some k ∈ N,

It means that,

P(k) : 2 + 4 + 6 + … + (2k) = k(k+1)

To prove that P(k + 1) is true, we have

2 + 4 + 6 + … + (2n) + 2(n + 1)

= k(k+1) + 2(k + 1)

= k2 + k + 2k + 2

= (k+1)(k+2)

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

Prove that 7n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 7n > n

If n =1, 71>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

7k > k …(1)

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

Now, multiplying both sides of the equation (1) by 7, we get

7 x 7k > 7k

Now by using the property,

i.e., 7k+1> 7k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

Prove that -1n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -1n > n

If n =1, -11<1. Hence P(1) is False

Therefore, P(n) isn’t  true for every positive integer n is proved using the principle of mathematical induction.

Prove that -1n <  n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): -1n < n

If n =1, -11<1. Hence P(1) is true

If n =2, -12=1. Hence P(2) isn’t  true

Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.

Prove that 1 + 2 + 3 + … + n  = n(n+1)/2 using the principle of Mathematical induction.

જવાબ : Given, 1 + 2 + 3 + … + n  = n(n+1)/2

Let, P(n) : 1 + 2 + 3 + … + n  = n(n+1)/2 , for n ∈ N

P(1) is true, as

P(1) : 1 = 1(1+1)/2

Let as assume P(k) is true for some k ∈ N,

It means that,

P(k) : k(k+1)/2

To prove that P(k + 1) is true, we have

1 + 2 + 3 + … + k + (k + 1)

= k(k+1)/2 + (2k + 1)

= (k2 + k + 4k + 2)/2

= (k + 1)(k + 2)/2

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

Prove that 13 + 23 + 33 + … + n3  = [n(n+1)/2]2 using the principle of Mathematical induction.

જવાબ : Given, 13 + 23 + 33 + … + n3  = [n(n+1)/2]2

Let, P(n) : 13 + 23 + 33 + … + n3  = [n(n+1)/2]2 , for n ∈ N

P(1) is true, as

P(1) : 1 = [1(1+1)/2]2

Let as assume P(k) is true for some k ∈ N,

It means that,

P(k) : [k(k+1)/2]2

To prove that P(k + 1) is true, we have

13 + 23 + 33 + … + k3 + (k+1)3

= [k(k+1)/2]2 + (k+1)3

= [(k+1)(k+2)/2]2

Hence, P(k + 1) is true, whenever P(k) is true.

Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

Prove that 3n > n for all positive integers n by the Principle of Mathematical Induction

જવાબ : Assume that P(n): 3n > n

If n =1, 31>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

3k > k …(1)

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

Now, multiplying both sides of the equation (1) by 3, we get

3 x 3k > 3k

Now by using the property,

i.e., 3k+1> 3k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.

### There are No Content Availble For this Chapter

 1 1.2 + 2.3 + 3.4 + ... + n.(n + 1) A n(4n2−1)/3 2 ½ + ¼ + 1/8 +...+ 1/2n B n(n+1)(n+2)/3 3 12 + 32 + 52 + ... + (2n − 1)2 C 1 – 1/2n

જવાબ :

1-B,2-C,3-A

 1 1.2 + 2.22 + 3.23 + ... + n.2n A n(n+1)(2n+7)/6 2 2 + 5 + 8 + 11 + ... + (3n − 1) B ½ n(3n+1) 3 1.3 + 2.4 + 3.5 + ... + n. (n + 2) C (n − 1) 2n+1+2

જવાબ :

1-C, 2-B, 3-A

 1 12.5+15.8+18.11+...+1(3n−1)(3n+2) A n/3n+1 2 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/(3n−2)(3n+1) B n/3(2n+3) 3 1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(2n+1)(2n+3) C n/3(4n+3) 4 1/3.7 + 1/7.11 + 1/11.5 +...+ 1/(4n−1)(4n+3) D n/(6n+4)

જવાબ :

1-D, 2-A, 3-B, 4-C

 1 7 + 77 + 777 + ... + 777  ...........n−digits7 A x + y 2 For all n ∈ N, x2n−1 + y2n−1 is divisible by B n2 3 For all n ∈ N, the sum of first n odd natural numbers is C 7(10n+1−9n−10)/81

જવાબ :

1-C, 2-A, 3-B

 1 12 + 32 + 52 + ... + (2n − 1)2 A 2n+1 – 1 2 22 + 42 + 62 + …+ 2n2 B 3 3 For all n ∈ N, n3 −- 7n + 3 is divisible by C [n(2n+1)(2n-1)]/3 4 For all n ∈ N, 1 + 2 + 22 + ... + 2n D n(4n2−1)/3

જવાબ :

1-D, 2-C, 3-B, 4-A

 1 For all n ∈ N, n(n + 1) (n + 5) is a multiple of A 133 2 For all n ∈ N, 72n + 23n−3. 3n−1 is divisible by B 24 3 For all n ∈ N, 2.7n + 3.5n − 5 is divisible by C 3 4 For all n ∈ N, 11n+2 + 122n+1 is divisible by D 25

જવાબ :

1-C, 2-D, 3-B, 4-A

 1 The sum of the series 1² + 2² + 3² + ………..n² is A n(n+1)/2 2 The sum of the series 1 + 2 + 3 + ………..n is B (n(n+1)/2)2 3 The sum of the series 13 + 23 + 33 + ………..n3 is C n(n+1)(2n+1)/6

જવાબ :

1-C, 2-A, 3-B

 1 For all n ∈ N, 52n+2 −24n −25 is divisible by A 2 2 For all n ∈ N, 32n+2 −8n − 9 is divisible by B 576 3 For all n ∈ N, n3 – n is divisible by C 8 4 For all n ∈ N, n2 – n is divisible by D 6

જવાબ :

1-B, 2-C, 3-D, 4-A

 1 If P (n) is the statement "n2 − n + 41, then P(1) is A 24 2 If P (n) is the statement "n2 − n + 41, then P(41) is B 8 3 For all n ∈ N, 52n −1 is divisible by C Prime 4 For all n ∈ N, 32n+7 is divisible by D Not Prime

જવાબ :

1-C, 2-D, 3-A, 4-B

 1 P(n) : 2n < n! A n ≥ 8 2 P(n) : 3n < n! B n ≥ 5 3 P(n) : 4n < n4 C n ≥ 3 4 P(n) : 7n < n7 D n ≥ 4

જવાબ :

1-D, 2-C, 3-B, 4-A

### Take a Test

Choose your Test :

Math

### Browse & Download CBSE Books For Class 11 All Subjects

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.