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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

 

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જવાબ : True


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

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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

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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

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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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : True


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

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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)

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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)

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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

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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.

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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.

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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

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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?

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જવાબ : False


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

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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

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જવાબ : True


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

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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.

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જવાબ : True


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

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જવાબ : True


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

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જવાબ : 1


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

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જવાબ : 1


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

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


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

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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)

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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)

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જવાબ : P(1) =12−1+41=41  (prime)


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

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જવાબ : 8


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

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જવાબ : 7


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

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જવાબ : 6


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

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જવાબ : 5


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

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જવાબ : 4


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

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જવાબ : 4


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

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જવાબ : 3


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

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જવાબ : 3


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

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જવાબ : 3


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

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જવાબ : 4


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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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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.

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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

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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.

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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

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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

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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

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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.

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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.

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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

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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

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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

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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)

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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

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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 −- 7+ 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

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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

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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

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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

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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

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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

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જવાબ :

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

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Mathematical Induction

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