Question

Prove the following by using the principle of mathematical induction for all n∈N: 1 + 3 + 3^2 + ... 3^(n-1) = (3^n -1) / 2

Answer 1

Given : 1 + 3 + 3² + ... 3⁽ⁿ⁻¹⁾ = (3ⁿ -1  ) / 2

To find : Prove by mathematical induction

Solution:

1 + 3 + 3² + ... 3⁽ⁿ⁻¹⁾ = (3ⁿ -1  ) / 2

This is an GP

a= 1  r = 3

Summation of GP for n terms  = a(rⁿ -1  ) / 2  =  (3ⁿ -1  ) / 2

But we have to prove by Mathematical induction

Here is proof :

use  n = 1

=> LHS = 1

     RHS = ( 3 - 1)/2  = 1

LHS =RHS

n = 2

=> LHS = 1 + 3 = 4

   RHS = ( 3² - 1)/2  = (9 - 1)/2 = 4

LHS = RHS

Lets assume its true for n = k

=> 1 + 3 + 3² + ... $3^{k-1}$ = ($3^k$ -1  ) / 2

Now lets check for n = k + 1

LHS  = 1 + 3 + 3² + ... $3^{k-1}$ + $3^k$  =  ($3^{k+1}$ -1  ) / 2

LHS = 1 + 3 + 3² + ... $3^{k-1}$ + $3^k$

= ($3^k$ -1  ) / 2 +  $3^k$

= ($3^k$ -1   + 2* $3^k$ ) / 2

= ( 3* $3^k$  - 1 ) / 2

=  ($3^{k+1}$ -1  ) / 2

= RHS

LHS = RHS

QED

Hence Proved

1 + 3 + 3² + ... 3⁽ⁿ⁻¹⁾ = (3ⁿ -1  ) / 2

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Answer 2

Answer:

Let the given statement be Perfect(n), i.e.,

P(n) : 1 + 3 + 3^2 + ..... + 3^n - 1 = (3^n - 1) /2

For n = 1,We have

P (1) : 1 = (3^1 - 1) /2 = 3 - 1/2 = 2/2 = 1,which is

true.

Let P(k) be true for some positive integer K,i.e.,

1 + 3 + 3^2 + .... + 3^k - 1 = (3k^ - 1) /2 -----> 1

We shall now prove that P(k + 1) is true.

Consider

1 + 3 + 3^2 +.... + 3^k - 1 + 3^(K + 1) - 1

= (1 + 3 + 3^2 + .... + 3^k -1 ) + 3^k

= (3^k - 1) /2 + 3^k [ Using Equation 1 ]

= (3^k - 1) + 2 . 3^k /2

= (1 + 2) 3^k - 1 /2

= 3.3^k - 1/2

= 3^k + 1 - 1 /2

Hence, Proved that

1 + 3 + 3^2 + ... + 3^n - 1 = (3^n - 1) /2

Step-by-step explanation:

@SSR