Latest TCS Aptitude Question SOLUTION: 1!+2!+3!+..........50!

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(#M40010539) TCS question series Keep an EYE

1!+2!+3!+..........50!

Asked In TCS rajat goel (11 years ago)

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abhilasha  (11 years ago)

plz explain in sinmple way..

Is this solution Helpfull? Yes (3) | No   

suman prabhakar  (11 years ago)

S = 3. something...3 x 10^64

Is this solution Helpfull? Yes (2) | No (3)  

PARAG KAMRA  (11 years ago)

multiply numerator and denominator by n(n+1)/2=50*51/2=25*51
thus u'll get (1*1!+2*2!+3*3!+4*4!+....+49*49!+50*50!)/(25*51)
=((2-1)*1!+(3-1)*2!+(4-1)*3!+.....+(51-1)*50!)/(25*51)
=(2!-1!+3!-2!+4!-3!+...+51!-50!)/(25*51)
=(51!-1!)/25*51
=2*49!-1/(25*51) [ans]
=> explanation
25*51=1275
so we take sum of 1275 like as
(1+2+3+..........+49)
and then multiply by each and we get
(1*1!+2*2!+3*3!+4*4!+....+49*49!+50*50!)
the answer of this ques after cal can be
3.04*10^64(ans)

Is this solution Helpfull? Yes (2) | No (9)  

ASWARTHA  (11 years ago)

ans is 3.1035*10^64

Is this solution Helpfull? Yes (2) | No (1)  

Jalesingh yadav  (11 years ago)

If the option is given my trick help to select option
its last digit is 3
because after the 5! last digit is zero so we add 1!+2!+3!+4! = 1+2+6+24 => 33
so the sum of 1!+2!+3!+.....50! gives last digit 3

Is this solution Helpfull? Yes (2) | No (1)  

akanchha gupta  (11 years ago)

giv me some proper solution for the problem

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