# 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