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Easy

Question

Find the sum open parentheses x plus 2 close parentheses to the power of n minus 1 end exponent plus open parentheses x plus 2 close parentheses to the power of n minus 2 end exponent open parentheses x plus 1 close parentheses plus open parentheses x plus 2 close parentheses to the power of n minus 3 end exponent open parentheses x plus 1 close parentheses to the power of 2 end exponent plus horizontal ellipsis plus open parentheses x plus 1 close parentheses to the power of n minus 1 end exponent

  1. open parentheses x plus 2 close parentheses to the power of n minus 2 end exponent minus open parentheses x plus 1 close parentheses to the power of n end exponent  
  2. open parentheses x plus 2 close parentheses to the power of n minus 1 end exponent minus open parentheses x plus 1 close parentheses to the power of n minus 1 end exponent  
  3. open parentheses x plus 2 close parentheses to the power of n end exponent minus open parentheses x plus 1 close parentheses to the power of n end exponent  
  4. None of these  

The correct answer is: open parentheses x plus 2 close parentheses to the power of n end exponent minus open parentheses x plus 1 close parentheses to the power of n end exponent


    We have,
    fraction numerator open parentheses x plus 2 close parentheses to the power of n end exponent minus open parentheses x plus 1 close parentheses to the power of n end exponent over denominator open parentheses x plus 2 close parentheses minus open parentheses x plus 1 close parentheses end fraction
    equals open parentheses x plus 2 close parentheses to the power of n minus 1 end exponent plus open parentheses x plus 2 close parentheses to the power of n minus 2 end exponent open parentheses x plus 1 close parentheses plus open parentheses x plus 2 close parentheses to the power of n minus 3 end exponent open parentheses x plus 1 close parentheses to the power of 2 end exponent plus horizontal ellipsis open parentheses x minus 1 close parentheses to the power of n minus 1 end exponent
    Hence, the required sum is
    open parentheses x plus 2 close parentheses to the power of n end exponent minus open parentheses x plus 1 close parentheses to the power of n end exponent blank left square bracket because blank open parentheses x plus 2 close parentheses minus open parentheses x plus 1 close parentheses equals 1 right square bracket

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