JMO VII and VIII -- series problems

If 

3! = 3 x 2 x 1;

4! = 4 x 3 x 2 x 1,

5! = 5 x 4 x 3 x 2 x 1

Then, find the value of 

1.1 ! + 2.2! + 3.3! + 4.4! + …… + 99.99!               

 JMO VII and VIII

Answer:

1.1! = (2 – 1).1! = 2x1 – 1.1!  = 2! – 1!

2.2! = (3 – 1).2! = 3x2! – 1.2! = 3! – 2!

3.3! = (4 – 1).3! = 4x3! – 1.3! = 4! – 3!

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99.99! = (100 – 1).99! = 100x99! – 1.99! = 100! – 99!

1.1 ! + 2.2! + 3.3! + 4.4! + …… + 99.99! = 100! – 1                   (adding all the terms)

Answer: 100! – 1