Question

To verify that the sum of the first n odd natural numbers is n2
can you give example in pattern

Answer 1

Answer:

Answer:

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =12

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = k

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)= k2 + 2k + 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)= k2 + 2k + 1= (k + 1)2

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)= k2 + 2k + 1= (k + 1)2P (n) is true for n = k + 1

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)= k2 + 2k + 1= (k + 1)2P (n) is true for n = k + 1Thus, P (n) is true for all n ∈ N.

Answer: Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n2Now, let us check for the P (n) is true for n = 1P (1) = 1 =121 = 1P (n) is true for n = 1Then, let’s us check for the P (n) is true for n = kP (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)Now, we have to show that1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2Then,1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1= k2 + (2k + 1)= k2 + 2k + 1= (k + 1)2P (n) is true for n = k + 1Thus, P (n) is true for all n ∈ N.