Maths-
General
Easy

Question

L t ┬ left parenthesis n rightwards arrow infinity right parenthesis left parenthesis 1 to the power of p plus 2 to the power of p plus 3 to the power of p plus midline horizontal ellipsis. plus n to the power of p right parenthesis divided by n to the power of left parenthesis p plus 1 right parenthesis is

  1. 1 divided by left parenthesis p plus 1 right parenthesis
  2. space 1 divided by left parenthesis 1 minus p right parenthesis
  3. 1 divided by p minus 1 divided by left parenthesis p minus 1 right parenthesis
  4. 1 divided by left parenthesis p plus 2 right parenthesis

hintHint:

In this question first we will rewrite the expression using begin inline style sum from r equals 1 to n of end style equals r to the power of 1 plus r squared plus r cubed plus. space. space. space. plus r to the power of n and n to the power of p plus 1 end exponent equals space n cross times n to the power of p .
We will assume r over n equals x then 1 over x equals d x then we will integrate the converted expression to find the limit.

The correct answer is: 1 divided by left parenthesis p plus 1 right parenthesis


    In this question we have to find the limit of limit as n rightwards arrow infinity of fraction numerator 1 to the power of p plus 2 to the power of p plus 3 to the power of p plus. space. space. space. plus n to the power of p over denominator n to the power of p plus 1 end exponent end fraction
    Step1: Rewriting the expression.
    We know that begin inline style sum from r equals 1 to n of end style equals r to the power of 1 plus r squared plus r cubed plus. space. space. space. plus r to the power of n and n to the power of p plus 1 end exponent equals space n cross times n to the power of p
    limit as n rightwards arrow infinity of fraction numerator sum from r equals 1 to n of r to the power of p over denominator n to the power of p cross times n end fraction
    => limit as n rightwards arrow infinity of sum from r equals 1 to n of open parentheses r to the power of p over n to the power of p close parentheses cross times 1 over n
    Step2: Finding the limit
    Let r over n equals x then 1 over x equals d x
    The expression converts to
    integral subscript 0 superscript 1 n to the power of p d x
    =>open square brackets fraction numerator n to the power of p plus 1 end exponent over denominator p plus 1 end fraction close square brackets subscript 0 superscript 1
    =>fraction numerator 1 over denominator p plus 1 end fraction
    Hence, the value of limit is fraction numerator 1 over denominator p plus 1 end fraction.

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