The product of the following series 1-1-1- - 1-2- - 1-3- - - - 1 - 1-1- - 1-2- - 1-3- - - is

The correct option is C 1 e^x=1+x+x^2/2!+x^3/3!...∞ Substituting x=1 , we get e^1=1+1+1/2!+1/3!...∞ ...(a)Substituting x= 1 , we get ...

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Standard XII

Mathematics

Logarithmic Differentiation

The product o...

Question

The product of the following series $(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $\times$ $(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . .)$ is

1

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e^{- 2}

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- e^{- 2}

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- 1

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Solution

The correct option is C $1$
$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} . . . \infty$
Substituting $x = 1$ , we get
$e^{1} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} . . . \infty$ ...(a)
Substituting $x = - 1$ , we get
$e^{1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} . . . \infty$ ...(b)
Therefore
$a \times b$
$= e^{1} (e^{- 1})$
$= 1$
Hence, option 'A' is correct.

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The product of the following series (1+11!+12!+13!+...) × (1−11!+12!−13!+...) is

The correct option is C 1ex=1+x+x22!+x33!...∞Substituting x=1, we gete1=1+1+12!+13!...∞ ...(a)Substituting x=−1, we gete1=1−1+12!−13!...∞ ...(b)Thereforea×b=e1(e−1)=1Hence, option 'A' is correct.

The product of the following series $(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $\times$ $(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . .)$ is