If $|\alpha|<1,|\beta|<1$ $1-\alpha+\alpha^2-\alpha^3+\ldots$. to $\infty=s_1$ $1-\beta+\beta^2-\beta^3+\ldots$ to $\infty=s_2$, then $1-\alpha \beta+\alpha^2 \beta^2-\alpha^3 \beta^3+\ldots$ to $\infty$ equals

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

|\alpha|<1,|\beta|<1

1-\alpha+\alpha^2-\alpha^3+\ldots

. to

\infty=s_1

1-\beta+\beta^2-\beta^3+\ldots

\infty=s_2

, then

1-\alpha \beta+\alpha^2 \beta^2-\alpha^3 \beta^3+\ldots

\infty

equals

View answer and explanation

Correct option: C

\frac{s_1 s_2}{1-s_1-s_2+2 s_1 s_2}

Option C (

\frac{s_1 s_2}{1-s_1-s_2+2 s_1 s_2}

) is correct because it correctly expresses the sum of the series in terms of

s_1

and

s_2

after substituting

\alpha

and

\beta

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}