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Question
CBSEENMA11012917

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

1 plus 3 plus 3 squared plus....... space plus 3 to the power of straight n minus 1 end exponent space equals space fraction numerator 3 to the power of straight n minus 1 over denominator 2 end fraction

Solution

 Let P(n) : WiredFaculty
I.    For n = 1,
      P(1) : 1 equals fraction numerator 3 to the power of 1 minus 1 over denominator 2 end fraction space rightwards double arrow space 1 space equals space 2 over 2 space rightwards double arrow space space space 1 space equals space 1

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

∴     P(m) : 1 plus 3 plus 3 squared plus space........... space plus space 3 to the power of straight m minus 1 end exponent space equals space fraction numerator 3 to the power of straight m minus 1 over denominator 2 end fraction     ... (i)

III.  
For   n = m + 1,                      
      P(m + 1) : WiredFaculty
or    straight P left parenthesis straight m plus 1 right parenthesis space colon space 1 plus 3 plus 3 squared plus.......... plus 3 to the power of straight m minus 1 end exponent space plus 3 to the power of straight m equals space fraction numerator 3 to the power of straight m minus 1 end exponent minus 1 over denominator 2 end fraction
  From (i), WiredFaculty

∴    straight P left parenthesis straight m space plus 1 right parenthesis space colon space open parentheses fraction numerator 3 to the power of straight m minus 1 over denominator 2 end fraction close parentheses space plus space 3 to the power of straight m space equals space fraction numerator 3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction
open parentheses fraction numerator 3 to the power of straight m minus 1 space plus space 2. space 3 to the power of straight m over denominator 2 end fraction close parentheses space equals space fraction numerator 3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction rightwards double arrow space space fraction numerator 3.3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction space equals space fraction numerator 3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction space space rightwards double arrow space fraction numerator 3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction space equals space fraction numerator 3 to the power of straight m plus 1 end exponent minus 1 over denominator 2 end fraction
which is true.

∴   P(m + 1) is true.

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




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