JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Hard

1 of 1

Analyzing +4 | -1

\log \left( {1 - x + {x^2}} \right) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \ldots ,

and

n

is not a multiple of 3 then

{a_n}

is equal to

View answer and explanation

Correct option: B

\frac{{{{\left( { - 1} \right)}^n}}}{n}

Option B (

\frac{{{{\left( { - 1} \right)}^n}}}{n}

) is correct because it represents the coefficient of

x^n

in the logarithmic expansion when n is not a multiple of 3.

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}