Sequence And Series Question 523
$ \frac{x^{2}-y^{2}}{1!}+\frac{x^{4}-y^{4}}{2!}+\frac{x^{6}-y^{6}}{3!}+……\infty = $
Options:
A) $ e^{x}-e^{y} $
B) $ {e^{x^{2}}}-{e^{y^{2}}} $
C) $ 2+{e^{x^{2}}}-{e^{y^{2}}} $
D) $ \frac{e^{x}-e^{y}}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ S=({e^{x^{2}}}-1)-({e^{y^{2}}}-1)={e^{x^{2}}}-{e^{y^{2}}} $ .
BETA
