Sets

Sets

Question

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N:

open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses space........ open parentheses 1 plus 1 over straight n close parentheses space equals space left parenthesis straight n space plus space 1 right parenthesis

Answer

Let P(n): open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses....... open parentheses 1 plus 1 over straight n close parentheses space equals space left parenthesis straight n plus 1 right parenthesis

I.       
For  n = 1,
        P(1) : open parentheses 1 plus 1 over 1 close parentheses space equals space 1 space plus space 1 space rightwards double arrow space 1 plus space 1 space equals space 1 plus space 1 space rightwards double arrow space 2 space equals space 2

∴       P(1) is true
II.      Suppose that the statement P (n) is true for n = m,  straight m space element of space straight N

∴      P(m) : open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses....... open parentheses 1 plus 1 over straight m close parentheses space equals space straight m plus space 1                 ...(i)
III.     For n = m + 1,
        straight P space left parenthesis straight m space plus space 1 right parenthesis colon space open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses........ space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses equals space straight m space plus space 1 space plus space 1
or      space space open square brackets open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses...... open parentheses 1 plus 1 over straight m close parentheses close square brackets space space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses space equals space straight m space plus space 2
       From (i), open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses space....... open parentheses 1 plus 1 over straight m close parentheses space equals space straight m space plus space 1

∴   straight P space left parenthesis straight m space plus space 1 right parenthesis colon space left parenthesis straight m space plus space 1 right parenthesis space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses space space equals space straight m space plus space 2 space space rightwards double arrow space left parenthesis straight m space plus space 1 right parenthesis space open parentheses fraction numerator straight m plus 1 plus 1 over denominator straight m plus 1 end fraction close parentheses space equals space straight m space plus space 2
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              which is true
∴             P (m + 1) is true

∴             P(m) is truerightwards double arrowP (m + 1) is true
Hence, by the principle of mathematical induction,  P(n) is true for all straight n element of space straight N.

         


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