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Numerical Ability
Number System
Q. The sum of (1*1!)+(2*2!)+(3*3!)+....+(1201*1201!)
Read Solution (Total 3)
-
- 1*1!=1 => 2!-1
1*1!+2*2!=1+4=5 => 3!-1
1*1!+2*2!+3*3!=1+4+18=23 => 4!-1
.....
........
similarly
(1*1!)+(2*2!)+(3*3!)+....+(1201*1201!)=(1202!-1)
- 11 years agoHelpfull: Yes(50) No(0)
- (1*1!)+(2*2!)+.....+(1201*1201!) can be written as (2!-1!)+(3!-2!)+(4!-3!)+....+(1202!-1201!).so, answer is 1202!-1
- 11 years agoHelpfull: Yes(3) No(0)
- we can write series like this (2!-1!)+(3!-2!)+(4!-3!)+....+(1202!-1201!)
now
all fro 2! to 1201!are cancel with each other
1202!-1! is left so
= 1202!-1 ans... - 11 years agoHelpfull: Yes(1) No(0)
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