Solution for the question - {:4t(pi rarr oo)(1^(rho)+2^(rho)+3^(rho)+cdots.+n^(rho))//pi^(()rho+1) is 1//(beta+2) '/> 1//rho-1//(rho-1) '/> 1//(1-rho) '/> 1//(beta+1) '/>

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

$1 / (p + 1)$
$1 / (1 - p)$
$1 / p - 1 / (p - 1)$
$1 / (p + 2)$

hint Hint:

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 $fraction numerator 1 over denominator p plus 1 end fraction$ .

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

$1 / (p + 1)$
$1 / (1 - p)$
$1 / p - 1 / (p - 1)$
$1 / (p + 2)$

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

Related Questions to study

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

$L i m left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x t a n 2 x minus 2 x t a n x right parenthesis divided by left parenthesis left parenthesis 1 minus c o s invisible function application 2 x right parenthesis squared right parenthesis equals$

$L i m left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x t a n 2 x minus 2 x t a n x right parenthesis divided by left parenthesis left parenthesis 1 minus c o s invisible function application 2 x right parenthesis squared right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x squared plus 5 x plus 3 right parenthesis divided by left parenthesis x squared plus x plus 2 right parenthesis right parenthesis to the power of x equals$

$L t subscript left parenthesis x rightwards arrow infinity right parenthesis invisible function application left parenthesis left parenthesis x squared plus 5 x plus 3 right parenthesis divided by left parenthesis x squared plus x plus 2 right parenthesis right parenthesis to the power of x equals$

$L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x plus a right parenthesis divided by left parenthesis x plus b right parenthesis right parenthesis to the power of left parenthesis x plus b right parenthesis equals$

$L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x plus a right parenthesis divided by left parenthesis x plus b right parenthesis right parenthesis to the power of left parenthesis x plus b right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis left parenthesis 1 minus e to the power of x right parenthesis s i n invisible function application x right parenthesis divided by left parenthesis x squared plus x cubed right parenthesis equals$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis left parenthesis 1 minus e to the power of x right parenthesis s i n invisible function application x right parenthesis divided by left parenthesis x squared plus x cubed right parenthesis equals$

With Turito Academy.

With Turito Foundation.

Get an Expert Advice From Turito.

Turito Academy

With Turito Academy.

Test Prep

With Turito Foundation.

Follow Us at

Support

Download App

Quick Links

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

is

In this question first we will rewrite the expression using and .We will assume then then we will integrate the converted expression to find the limit.

The correct answer is:

In this question we have to find the limit of Step1: Rewriting the expression.We know that and => Step2: Finding the limitLet then The expression converts to=>=>Hence, the value of limit is .

Related Questions to study

For

For

, given that and

, given that and

If then the value of is,

If then the value of is,

If then has the value

If then has the value

If then is

If then is

If then

If then

where a,b,c are real and non-zero=

where a,b,c are real and non-zero=

With Turito Academy.

With Turito Foundation.

Get an Expert Advice From Turito.

Turito Academy

With Turito Academy.

Test Prep

With Turito Foundation.

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

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$