Solution for the question - 4(pi rarr oo)[1//n^(2)sec^(2)1//n^(2)+2//n^(2)sec^(2)4//n^(2)+cdots++1//nsec 1]= 1//2tan 1 '/> tan 1 '/> 1//2cos ec1 '/> 1//2 secl '/>

Maths-

General

Easy

Question

$L t left parenthesis n rightwards arrow infinity right parenthesis left square bracket 1 divided by n squared s e c squared 1 divided by n squared plus 2 divided by n squared s e c squared 4 divided by n squared plus midline horizontal ellipsis. plus 1 divided by n s e c squared 1 right square bracket equals$

$1 / 2 s e c 1$
$1 / 2 \cos e c 1$
$\tan 1$
$1 / 2 \tan 1$

hint Hint:

In this question we have use the formula of the sum $open square brackets 1 over n squared sec squared open parentheses 1 over n squared close parentheses plus 2 over n squared sec squared open parentheses 4 over n squared close parentheses plus... plus 1 over n squared sec squared open parentheses 1 close parentheses close square brackets space equals space begin inline style sum from r equals 1 to r equals n of end style open parentheses r over n squared close parentheses sec squared open parentheses r over n close parentheses squared$
Then we assume $r over n equals x$ then $1 over n equals space d x$ and then we integrate the expression with proper limits to get the answer.

The correct answer is: $1 divided by 2 space tan invisible function application 1$

In this question we have to find the value of $limit as n rightwards arrow infinity of open square brackets 1 over n squared sec squared open parentheses 1 over n squared close parentheses plus 2 over n squared sec squared open parentheses 4 over n squared close parentheses plus... plus 1 over n squared sec squared open parentheses 1 close parentheses close square brackets$
Step1: Rewriting the expression
We know that
$open square brackets 1 over n squared sec squared open parentheses 1 over n squared close parentheses plus 2 over n squared sec squared open parentheses 4 over n squared close parentheses plus... plus 1 over n squared sec squared open parentheses 1 close parentheses close square brackets space equals space begin inline style sum from r equals 1 to r equals n of end style open parentheses r over n squared close parentheses sec squared open parentheses r over n close parentheses squared$

Step2: Finding the limit
$limit as n rightwards arrow infinity of sum from r equals 1 to r equals n of open parentheses r over n squared close parentheses sec squared open parentheses r over n close parentheses squared$
Let $r over n equals x$ then $1 over n equals space d x$
Then $integral subscript 0 superscript 1 x sec squared open parentheses x squared close parentheses d x$
By putting $x squared equals t$ we get,
$1 half integral subscript 0 superscript 1 sec squared open parentheses t close parentheses d t$
By Integrating the expression we get
=> $1 half left square bracket tan open parentheses t close parentheses right square bracket subscript 0 superscript 1$
=> $1 half tan open parentheses 1 close parentheses$

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$1 / 2 s e c 1$
$1 / 2 \cos e c 1$
$\tan 1$
$1 / 2 \tan 1$

The correct answer is: $1 divided by 2 space tan invisible function application 1$

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Can’t find what you’re looking for?

In this question we have use the formula of the sum Then we assume then and then we integrate the expression with proper limits to get the answer.

The correct answer is:

In this question we have to find the value of Step1: Rewriting the expressionWe know thatStep2: Finding the limitLet then Then By putting we get,By Integrating the expression we get=>=>

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The correct answer is: $1 divided by 2 space tan invisible function application 1$

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