Latest TCS Aptitude Question SOLUTION: 1! + 2! + .... 50! = ? a. 3.1035*10^64 b. 2.1021* 10 ^ 65 c. 3.1035*10^63 d. 3.1035*10^62

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(#M40001195) TCS question Keep an EYE

1! + 2! + .... 50! = ?

a. 3.1035*10^64
b. 2.1021* 10 ^ 65
c. 3.1035*10^63
d. 3.1035*10^62

Asked In TCS ranjani (13 years ago)

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khushboo mishra  (13 years ago)

as 50!=3.0414093202×10^64
refer:-http://en.wikipedia.org/wiki/Factorial
so by the below mentioned approach option a is correct....

ans:-(a)

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khushboo mishra  (13 years ago)

this summation can be defined by reverse recurssion
if you want the sum of the factorials from 1 to n then start with
s(n) = n
and then
s(i) = i * ( 1 + s(i+1) )
and s(1) gives the summation
for example if n=4 we have
s(4)=4
s(3)=3*(4+1)=15
s(2)=2*(15+1)=32
s(1)=1*(32+1)=33
which agrees with
1!+2!+3!+4! = 1+2+6+24

using a computer program and using this recurssion method I get the sum as
31035053229546199656252032972759319953…

this is wat i can help...

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