Question

1. Simplify 1/2! + 2/3! + 3/4! ... + 11/12!
2.let x=2^100,y^60 and z=10^30.What is the smallest number among the three?


can you check my answers?
1. 1
2. y=3^60

Answer 1

#1
sum(1) = 1/2
sum(2) = 1/2 + 2/3!
= 1/2 + 1/3 = 5/6
sum(3) = 5/6 + 3/4! = ( (3+1)! - )/(3+1)!
= 5/6 + 1/8 = 23/24
sum(4) = 23/24 + 4/5!
= 23/24 + 1/30 = 119/120 = ((4+1)! - 1)/(4+1)!
it looks like
sum(n) = ( (n+1)! - 1)/(n+1)!

so sum(11) = (12! - 1)/12! = 479001599/479001600 which is not yet 1, but sure is close to 1

#2 is correct, did you use your calculator?

Answer 2

My solution is too long. I didn't use calculator

Answer 3

for #2, I suggest taking logs
x = 2^100
logx = 100log2 = 30.102..
y = 3^60
logy = 60log3 = 28.627..
z = 10^30
logz = 30log10 = 30

thus logy < logz < logx
then y < z < x ----> y is the smallest