If $$A = \sum\limits_{n = 1}^\infty {\frac{{2n}}{{\left( {2n - 1} \right)!}},B = \sum\limits_{n = 1}^\infty {\frac{{2n}}{{\left( {2n + 1} \right)!}}} } $$ then $$AB$$ is equal to
A.
$$e^2$$
B.
$$e$$
C.
$$e + e^2$$
D.
$$1$$
Answer :
$$1$$
Solution :
$$\eqalign{
& A = \sum\limits_{n = 1}^\infty {\frac{{2n - 1 + 1}}{{\left( {2n - 1} \right)!}}} \cr
& = \sum\limits_{n = 1}^\infty {\left[ {\frac{1}{{\left( {2n - 2} \right)!}} + \frac{1}{{\left( {2n - 1} \right)!}}} \right]} \cr
& = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ..... = e \cr} $$
Similarly $$B = e^{- 1}$$ as terms will be alternately positive and negative.
$$\therefore AB = e \cdot {e^{ - 1}} = 1$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is