Using the principle of mathematical induction, prove each of the following for all n ϵ N: 1 + 3 + 3^2 + 3^2 +......+ 3^(n-1) = 1/2(3^n - 1) ​

Question

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

1 + 3 + 3² + 3² +......+ 3^n-1 = \(\frac{1}2\)(3ⁿ - 1)

Answer 1

To prove:

1 + 3¹ + 3² +......+ 3^n-1 = \(\frac{3^n-1}2\)

Steps to prove by mathematical induction: 

Let P(n) be a statement involving the natural number n such that 

(i) P(1) is true 

(ii) P(k + 1) is true, whenever P(k) is true 

Then P(n) is true for all n ϵ N 

Therefore,

Let P(n): 1 + 3¹ + 3² +......+ 3^n-1 = \(\frac{3^n-1}2\)

Step 1:

P(1) = \(\frac{3^1-1}2\) = \(\frac{2}2\) = 1

Therefore, P(1) is true 

Step 2: 

Let P(k) is true Then,

p(k): 1 + 3¹ + 3² +......+ 3^k-1 = \(\frac{3^k-1}2\)

Now,

 1 + 3¹ + 3² +......+ 3^k-1 + 3^{(k +1)-1} = \(\frac{3^{(k)}-1}2\) + 3^{(k + 1)-1}

= P(k + 1)

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

Hence, by the principle of mathematical induction, we have

1 + 3¹ + 3² + ....... + 3^n-1 = \(\frac{3^n-1}2\) for all n ϵ N