The expansion of e-e-122-e-e-122 is

The correct option is B 1 Using the identity (a^2 b^2)=(a+b)(a b) =(e+e^ 1 e+e^ 12)(e+e^ 1+e e^ 12) =(e^ 1)(e) =1 ...

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Progression

The expansion...

Question

The expansion of ${(\frac{e + e^{- 1}}{2})}^{2} - {(\frac{e - e^{- 1}}{2})}^{2}$ is

1

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2

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3

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4

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E = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . .

\frac{1}{2 +} \frac{2}{3 +} \frac{3}{4 +} \dots = \frac{3 - e}{e - 2}

e

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e^{2} - e = 1 + \frac{1 + 2}{a!} + \frac{1 + 2 + 2^{2}}{3!} + \frac{1 + 2 + 2^{2} + 2^{3}}{c!} + . . .

c \div a

E_{1}, E_{2}, E_{3} a n d E_{4}

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4^{t h}

E_{1}, E_{2}, E_{3} a n d E_{4}

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