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Q. The value of $ \frac{1}{2!}+\frac{2}{3!}+....+\frac{999}{1000!} $ is equal to

KEAMKEAM 2009

Solution:

$ \frac{1}{2!}+\frac{2}{3!}+....+\frac{999}{1000!} $
$=\frac{2-1}{2!}+\frac{3-1}{3!}+.....+\frac{1000-1}{1000!} $
$=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{999!}-\frac{1}{1000!} $
$=1-\frac{1}{1000!}=\frac{1000!-1}{1000!} $