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Sets

Question
CBSEENMA11012921

Prove the following by using the principle of mathematical induction for allspace space space space straight n element of straight N.

WiredFaculty.

Solution

Let P(n): WiredFaculty
I.     For n =1,
      WiredFaculty
∴     P(1) is true.
II.   Suppose the statement is true for n = m, straight m space element of space straight N.

∴    straight P left parenthesis straight m right parenthesis colon space space 1.2.3 space plus space 2.3.4 space plus space 3.4.5 space plus space......... space plus space straight m space left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space equals space fraction numerator straight m space left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis over denominator 4 end fraction space space... space left parenthesis straight i right parenthesis
III.   For n = m + 1,
        WiredFaculty
                                                                         equals fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis left parenthesis straight m plus 3 right parenthesis left parenthesis straight m plus 4 right parenthesis over denominator 4 end fraction
or     1.2.3 + 2.3.4 + 3.4.5 + ........... + m (m + 1) (m + 2) + (m + 1) (m + 2) (m + 3)
                                                                           equals fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 end fraction
        From (i), 1.2.3 + 2.3.4 + 3.4.5 + ......... + m (m + 1) (m + 2)
                                                                        WiredFaculty

∴    WiredFaculty
                                                                           equals fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 end fraction
rightwards double arrow space space space space space space space space space left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis space open square brackets straight m over 4 plus 1 close square brackets space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 end fraction
rightwards double arrow        WiredFaculty
             which is true.

∴        P(m + 1) is true.

∴        P(m) is true rightwards 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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