Prove by the principle of mathematical induction: 1 + 3 + 32 + … +

1.	Prove by the principle of mathematical induction: 1 + 3 + 32 + … + 3n-1 = (3n – 1)/2
Answer» Suppose P (n) = 1 + 3 + 3² + – – – – + 3^{n – 1} = (3ⁿ – 1)/2 Then, For n = 1 P (1) = 1 = (3¹ – 1)/2 = 2/2 =1 P (n) is true for n = 1 Then, let’s us check for the P (n) is true for n = k P (k) = 1 + 3 + 3² + – – – – + 3^{k – 1} = (3^k – 1)/2 … (i) Then, we have to show P (n) is true for n = k + 1 P (k + 1) = 1 + 3 + 3² + – – – – + 3^k = (3^{k + 1} – 1)/2 Now, {1 + 3 + 3² + – – – – + 3^{k – 1}} + 3^{k + 1 – 1} = (3k – 1)/2 + 3^k using equation (i) = (3k – 1 + 2 × 3^k)/2 = (3 × 3 k – 1)/2 = (3^{k + 1} – 1)/2 P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction: 1 + 3 + 32 + … + 3n-1 = (3n – 1)/2

Answer»

Suppose P (n) = 1 + 3 + 3² + – – – – + 3^{n – 1} = (3ⁿ – 1)/2

Then, For n = 1

P (1) = 1 = (3¹ – 1)/2 = 2/2 =1

P (n) is true for n = 1

Then, let’s us check for the P (n) is true for n = k

P (k) = 1 + 3 + 3² + – – – – + 3^{k – 1} = (3^k – 1)/2 … (i)

Then, we have to show P (n) is true for n = k + 1

P (k + 1) = 1 + 3 + 3² + – – – – + 3^k = (3^{k + 1} – 1)/2

Now, {1 + 3 + 3² + – – – – + 3^{k – 1}} + 3^{k + 1 – 1}

= (3k – 1)/2 + 3^k using equation (i)

= (3k – 1 + 2 × 3^k)/2

= (3 × 3 k – 1)/2

= (3^{k + 1} – 1)/2

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.