JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots

upto

\infty=\frac{\pi^2}{6}

, then value of

\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots

upto

\infty

View answer and explanation

Correct option: C

\frac{\pi^2}{8}

Option C (

\frac{\pi^2}{8}

) is correct because it is derived by subtracting the sum of even-numbered terms from the total sum of the series.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

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

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

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

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

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

y

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

y=x-2

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

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

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

A P

a, b

G P

a, b, c

\frac{3}{4}