Sets

Sets

Question

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

1 cubed plus 2 cubed plus 3 cubed plus space... space plus space straight n cubed space equals space open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets squared

Answer

Let straight P left parenthesis straight n right parenthesis space colon space 1 cubed plus 2 cubed plus 3 cubed plus..... plus straight n cubed space equals space open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets squared
I.      For n = 1,
       straight P left parenthesis 1 right parenthesis colon space 1 cubed space equals space open square brackets fraction numerator 1 left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction close square brackets squared space rightwards double arrow 1 space equals space 1 space rightwards double arrow space straight P left parenthesis 1 right parenthesis is true.
II.    Suppose the statement is true for n = m, straight m space element of space straight N
   
          i.e., "<pre    ... (i)
III.     For n = m + 1,
        straight P left parenthesis straight m plus 1 right parenthesis colon space 1 cubed plus 2 cubed plus 3 cubed plus......... plus left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
or     left square bracket 1 cubed space plus space 2 cubed space plus space 3 cubed space plus space....... space plus space straight m cubed right square bracket space plus space left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
     From (i), 1 cubed plus 2 cubed plus 3 cubed plus space............ space plus space straight m cubed space equals space open square brackets fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction close square brackets squared

∴   space space straight P left parenthesis straight m plus 1 right parenthesis space colon space open square brackets fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction close square brackets squared space plus space left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
rightwards double arrow space space left parenthesis straight m plus 1 right parenthesis squared space open square brackets straight m squared over 4 plus left parenthesis straight m plus 1 right parenthesis close square brackets space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
rightwards double arrow space space space left parenthesis straight m plus 1 right parenthesis squared open parentheses fraction numerator straight m squared plus 4 straight m plus 4 over denominator 4 end fraction close parentheses space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
rightwards double arrow space space space fraction numerator left parenthesis straight m plus 1 right parenthesis squared left parenthesis straight m plus 2 right parenthesis squared over denominator 4 end fraction space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared space rightwards double arrow space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared
       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 straight n element of space straight N.



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