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

• Placement Papers
• /
• TCS
• /
• Logical Reasoning
• /
• Number Series

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40000397) TCS Logical Reasoning question Keep an EYE

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

Solution

Not possible to evaluate exactly if ! is the sign for factorial.

Like manish sir answer on facebook.

Asked In TCS V.V.Ravi Teja (13 years ago)

Solved Read Solution (7)

Is this Puzzle helpful?   (93)   (34) Submit Your Solution Number Series

Read Solution (7) :

ravichandra  (9 years ago)

Go for options --> Units digit must be "3"

because
1! = 1 (units digit 1)
2! = 2 (units digit 2)
3! = 6 (units digit 6)
4! = 24 (units digit 4)
5!, 6!.... the units digit will be 0 because all the terms will include 5x2=10

units digit of 1!+2!+... will be 1+2+6+4+0+0+0.... = 3

Is this solution Helpfull? Yes (45) | No (5)  

Manuj Agarwal  (13 years ago)

ans is 3*10^64

Is this solution Helpfull? Yes (13) | No (10)  

Chetan Anand  (12 years ago)

3.103505323*10^64

Is this solution Helpfull? Yes (11) | No (5)  

hemant panchal  (10 years ago)

In this series the nth term is 50
so, (n + 1)!/n = 51!/50
But this expression will give the closest possible value to the exact answer.

Is this solution Helpfull? Yes (11) | No (11)  

chintan Makwana  (12 years ago)

I believe there is no simple expression for this number!

Let's try approximating such a sum for more terms! This ultimately needs Stirlings' approximation for n!

wolframalpha can handle the sizes of the 50 factorials involved and their sum, so the sum can be calculated exactly using it (with a lot of work).

Call the sum S, then 50! < S < 2*50! This is because 50! = 50*49! = 49! + 49! + 49! + ... +49!, which has 50 terms, each as large as the 49 other terms in 1!+2!+3!.......49!

30 414 093 201 713 378 043 612 608 166 064 768 844 377 641 568 960 512 000 000 000 000 < S <
60 828 186 403 426 756 087 225 216 332 129 537 688 755 283 137 921 024 000 000 000 000

It factors as (2^E)(3^F)(5^G)*K, with E=0, F=1, G=0

The least significant digits can be calculated, because they remain fixed! Starting at 1!+2!+3!+4! = 33, the units digit will remain 3 indefinitely. The sum at 9! will have the fixed last two digits (10! ends with two 0s).

The most signigicant digits can be found in a similar way, pretty much done above, with 50! having the same leading digit as S. So
S = 3. something...3 x 10^64

Is this solution Helpfull? Yes (8) | No (19)  

indrani banerjee  (9 years ago)

50! * 51!/2 = answer as n(n+1)/2 is the formula fr summation.

Is this solution Helpfull? Yes (8) | No (14)  

musku rock krishna reddy  (8 years ago)

add first and last digit then second an last second digit continue this process .50+1*49+2*48+3...... and so on we get 25 51s the ans is 25*51=1275.

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.