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(#M40030008) TCS question Keep an EYE

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

Asked In TCS (11 years ago)

Unsolved Read Solution (6)

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DIVYA KUKRETI  (11 years ago)

ans: 50!(51)/50
Write eq as:
50!+49!+48!+...1!
50!(1+1/50+1/50*49....)
50!(1+1/50)... since other terms will be neglegible
50!(51)/50

Is this solution Helpfull? Yes (32) | No (3)  

Raushan shukla  (11 years ago)

for these type of exprsn formula z (n+1)!/n
so ans= 51!/50.

Is this solution Helpfull? Yes (14) | No (3)  

ARUN GUPTA  (11 years ago)

50!(!1/!50+!2/!50+!3/!50+......+!49/!50+!50/!50)
= !50(0+0+0+.......0+0+1/50+1)
= !50(1+1/50)
= !50*51/50

Is this solution Helpfull? Yes (7) | No (2)  

Nikhil Bansal  (11 years ago)

I think the options also play an important role and if they have values instead of factorial notation than best way to go about it is to find unit digit. Its very simple since we just have to calculate till 4!; reason is that from 5! onward each term will have a zero at its unit place. 1 + 2 + 6 + 4 + 0 + ..... + 0 = 3.

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

divya  (11 years ago)

It can be solved by : (n+1)!/n
so here answer will be : (50+1)!/50 = 51!/50.
If any query you can ask

Is this solution Helpfull? Yes (2) | No   

sumon maji  (11 years ago)

[{50*(50+1)}/2]!=25*51!=1275!

Is this solution Helpfull? Yes | No   

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