JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

Let

{S_n} = 1.(n - 1) + 2.(n - 2) + 3.(n - 3) + ..... + (n - 1).1,n \ge 4.

The sum

\sum\limits_{n = 4}^\infty {\left( {\frac{{2{S_n}}}{{n!}} - \frac{1}{{(n - 2)!}}} \right)}

is equal to

View answer and explanation

Correct option: A

\frac{{e - 1}}{3}

Option A (

\frac{{e - 1}}{3}

) is correct because it is the result of evaluating the given infinite sum by simplifying the expression and using the series expansion of e.

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}