Principle Of Mathematical Induction

Question
CBSEENMA11015574

Statement − 1: For every natural number n ≥ 2 fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction space greater than space square root of straight n

Statement −2: For every natural number n ≥ 2,straight n greater or equal than 2 comma space square root of straight n left parenthesis straight n plus 1 right parenthesis space end root space less than space straight n plus 1

  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement − 1 is true, Statement − 2 is false. 

Solution

C.

Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

straight P space left parenthesis straight n right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction
straight P space left parenthesis 2 right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space greater than space square root of 2
Let space us space assume space that space straight P space left parenthesis straight k right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space...... fraction numerator 1 over denominator square root of straight k end fraction space plus fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space greater than square root of straight k plus 1 end root
has space to space be space true.
straight L. straight H. straight S greater than thin space square root of straight k space plus space fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space equals space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root space plus 1 over denominator square root of straight k plus 1 end root end fraction
since space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space greater than straight k space space left parenthesis for all space straight k greater or equal than 0 right parenthesis
therefore space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root plus 1 over denominator square root of straight k plus 1 end root end fraction space greater than space fraction numerator straight k plus 1 over denominator square root of straight k plus 1 end root end fraction space equals space square root of straight k plus 1 end root
Let space straight p left parenthesis straight n right parenthesis space space equals space square root of straight n space left parenthesis straight n plus 1 right parenthesis end root space less than space straight n plus 1
State space minus 1 space is space correct.
straight P space left parenthesis 2 right parenthesis space space equals space square root of 2 space straight x space 3 end root space less than space 3
If space straight P space left parenthesis straight k right parenthesis space space equals space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space less than space left parenthesis straight k plus 1 right parenthesis space is space true
Now space space straight P space left parenthesis straight k plus 1 right parenthesis space equals space square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space less than space straight k plus 2 space has space to space be space true
square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space thin space left parenthesis straight k plus 2 right parenthesis
Hence Statement −2 is not a correct explanation of Statement −1. 

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