Sets

Sets

Question

Use principle of mathematical induction to prove that:

1 space plus space 2 space plus space 3 space plus space... space plus space straight n space equals space fraction numerator straight n left parenthesis straight n space plus space 1 right parenthesis over denominator 2 end fraction

Answer

Let P(n): 1 + 2 + 3 + ......... + n = space space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction
I. For n = 1,
    P(1) : 1 = fraction numerator 1 left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction rightwards double arrow space 1 space equals space 1 space rightwards double arrow space space straight P left parenthesis 1 right parenthesis is true.
II.  Suppose the statement is true for n = m, straight m element of straight N
      i.e. P(m): 1 plus 2 plus 3 plus........ space plus straight m space equals space fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction          ....(i)
III.    For n = m + 1,
        P(m + 1): 1 + 2 + 3 + ........ + (m + 1) = fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
or  [1 + 2 + 3 + ...... + m] + (m + 1) = fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
                                       
                                         [From (i), 1 + 2 + 3 + ...... + m = fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction]
∴        P (m + 1): space fraction numerator straight m left parenthesis straight m space plus space 1 right parenthesis over denominator 2 end fraction space plus space left parenthesis straight m space plus space 1 right parenthesis space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
rightwards double arrowspace space left parenthesis straight m plus 1 right parenthesis open parentheses straight m over 2 plus 1 close parentheses space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
rightwards double arrow   left parenthesis straight m plus 1 right parenthesis open parentheses fraction numerator straight m plus 2 over denominator 2 end fraction close parentheses space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
rightwards double arrowfraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
    which is true

∴    P(m + 1) is true

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


      

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