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  • 1 + 2 + 22 + ... 2n = 2n+1 -1 for all natural numbers n.

1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1 for all natural numbers n.

Answers (1)

P(n) is 1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1 ..................given

Now, we’ll substitute different values for n,

P(0) = 1 = 2^{0+1} -1 , is true

P(1) = 1 + 2 = 3 = 2^{1+1} -1 , is true

P(2) = 1 + 2+ 2^2 = 7 = 2^{2+1} -1, is true

Now, let us consider,

P(k) = 1 + 2 + 2^2 + .... 2^k = 2^{k+1} -1 to be true

Thus,

P(k+1) is 1+2+2^2+....+2^k + 2^{k+1 }= 2^{k+1} -1 + 2^{k+1}

            = 2 \times 2^{k+1 }-1

           = 2^{k+1+1 }-1

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) is 1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1.

 

Posted by

infoexpert21

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