Mathematics
Functions
Easy

Question

text  Let  end text f left parenthesis x right parenthesis equals x plus f left parenthesis x minus 1 right parenthesis text  for  end text straight for all x element of R. text  If  end text f left parenthesis 0 right parenthesis equals 1 comma text  find  end text f left parenthesis 100 right parenthesis

  1. 5051
  2. 5052
  3. 5053
  4. 5054

The correct answer is: 5051


    text  Given  end text f left parenthesis x right parenthesis equals x plus f left parenthesis x minus 1 right parenthesis text  and  end text f left parenthesis 0 right parenthesis equals 1
text  Put  end text space of 1em x equals 1. text  Then  end text
f left parenthesis 1 right parenthesis equals 1 plus f left parenthesis 0 right parenthesis equals 2
text  Put  end text space of 1em x equals 2 text  . Then,  end text
f left parenthesis 2 right parenthesis equals 2 plus f left parenthesis 1 right parenthesis equals 4
text  Put  end text space of 1em x equals 3. text  Then  end text
f left parenthesis 3 right parenthesis equals 3 plus f left parenthesis 2 right parenthesis equals 7
f left parenthesis 3 right parenthesis equals 3 plus f left parenthesis 2 right parenthesis equals 7
text  Thus,  end text f left parenthesis 0 right parenthesis comma f left parenthesis 1 right parenthesis comma f left parenthesis 2 right parenthesis comma horizontal ellipsis text  form a series  end text 1 comma 2 comma 4 comma 7 comma horizontal ellipsis
text  Let  end text space of 1em S equals 1 plus 2 plus 4 plus 7 plus midline horizontal ellipsis plus f left parenthesis n minus 1 right parenthesis
S equals 1 plus 2 plus 4 plus midline horizontal ellipsis plus f left parenthesis n minus 2 right parenthesis plus f left parenthesis n minus 1 right parenthesis
text  Subtracting, we get  end text
0 equals left parenthesis 1 plus 1 plus 2 plus 3 plus midline horizontal ellipsis plus n text  terms  end text right parenthesis minus f left parenthesis n minus 1 right parenthesis
therefore space of 1em f left parenthesis n minus 1 right parenthesis equals fraction numerator n left parenthesis n plus 1 right parenthesis over denominator 2 end fraction
therefore space of 1em f left parenthesis 100 right parenthesis equals 5051

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