JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Sum of 36 terms of the series

\frac{1}{9 \sqrt{11}+11 \sqrt{9}}+\frac{1}{11 \sqrt{13}+13 \sqrt{11}}+\frac{1}{13 \sqrt{15}+15 \sqrt{13}}+\ldots

is equal to

View answer and explanation

Correct option: B

\frac{1}{9}

Option B (

\frac{1}{9}

) is correct because by rationalizing the denominator and simplifying the series, it telescopes, leaving only the first and last terms, which allows us to calculate the sum.

More Questions from SEQUENCE OF SERIES

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Given the sets

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

, then find the following
(i)

{A} \cup({B} \cap {C})

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

{x^2},{b^2},{y^2}

are in Suppose

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

is expressed as a polynomial in

y

, as

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

where

y=x-2

, then

\sum\limits_{{i=0}}^{2012} a_i

equals If

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

and

\log \left(\frac{a}{3 b}\right)

are in

A P

, where

a, b

and c are in

G P

, then

a, b, c

are the lengths of sides of If the sum of three numbers in G.P. is 63 and the product of the first and the second term is

\frac{3}{4}

of the third term, then the numbers are If the lengths of sides of a right angled triangle are in A.P., then their ratio is