Question

Pease HELPP ME--------------------------------------USING MATEMATICAL INDUCTION PROVE THAT FOR ANY NATURAL NUMBER 'N' THE STATEMENT 4 2N>

Answer 1

the question is not clear for me. can you post help me in understanding the question

Answer 2

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Let’s call the given formula, S(n), and let it be a function of n as follows:

S(n) = 1 + 2 + 2² + ... + 2^(n ‒1) = 2^(n) ‒ 1 

Now, using the principle of mathematical induction to prove that the given formula S(n) is true for all natural numbers n, we’ll proceed as follows:

A.) Prove that the given formula S(n) is true for n = 1.
S(1) = 1 = 2¹ ‒ 1
           1 = 2 ‒ 1
           1 = 1
         Therefore, the given formula is true for n = 1.

B.) Now, prove that if the given formula S(n) is true for n = k, where k ≥ 1, then it is also true for n = k + 1.

First, assume that the given formula is true for n = k, and, consequently, we substitute into the given formula as follows:  

S(k) = 1 + 2 + 2² + … + 2^(k ‒ 1) = 2^(k) ‒ 1          (1)

Now, prove that the given formula it is true for n = k + 1.

    S(k + 1) = 1+ 2 + 2² + … + 2^(k ‒ 1) + 2^[(k + 1) ‒ 1] =  2^(k + 1) ‒ 1
                     1 + 2 + 2² + … + 2^(k ‒ 1) + 2^k  =  2^(k + 1) ‒1         (2)

But, from Part B above, we have equation (1): 
 
1 + 2 + 2² + … + 2^(k ‒ 1) = 2^(k) ‒ 1.  Substituting this result into the left side of equation (2), we get:

     2^(k) ‒ 1 + 2^k  = 2^(k + 1) ‒1

NOTE: We can make this substitution because we assumed that the given formula is true for n = k; therefore, we can treat the “n = k” case as a fact temporarily to see if it means that the “n = k + 1” case is also true.

Now, collecting like terms on the left side, we get:

     2[2^(k)] ‒ 1 = 2^(k + 1) ‒ 1

Since a = a¹ for any real number a, we have:

     (2¹)[2^(k)] ‒ 1= 2^(k + 1) ‒ 1

Now, by the Product of Two Powers property of positive integral exponents m and n, i.e., aᵐ(aⁿ) = a^(m + n), we have the following desired result: 

      2^(k + 1) ‒ 1 = 2^(k + 1) ‒ 1

Since both sides of the equation are now shown to be the same, the equation is therefore true (satisfied), thus showing that the given formula is true for n = k + 1,thus, in turn, showing that when the given formula is true for n = k, it is also true for n = k + 1, and, therefore it is true for all natural numbers (positive integers) n.

Some examples illustrating the veracity of the above proof of the given formula:

S(1) = 1 = 2¹ ‒ 1
           1 = 2 ‒ 1
           1 = 1   (shown previously)

S(2) = 1 + 2 = 2² ‒ 1
                  3 = 3

S(3) = 1 + 2 + 2² = 2³ ‒ 1
            1 + 2 + 4 = 8 ‒ 1
                         7 = 7

S(4) = 1 + 2 + 2² + 2³ = 2^4 ‒ 1
           1 + 2 + 4 + 8 = 16 ‒ 1
                            15 = 15    
           
S(5) = 1 + 2 + 2² + 2³ + 2^4 = 2^5 ‒ 1
               1 + 2 + 4 + 8 + 16 = 32 ‒ 1
                                        31 = 31


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