Question #42814

1 Answer
Dec 20, 2017

The intrest will be #2*10^6(e^0.02-1)~=40402.68# dollars.

Explanation:

If the account pays #2%# half-yearly, the interest after one year is #2000000*{(1+0.02/2)^2-1}=40200# dollars.

And, if the account pays the interest monthly, the interest he will have after a year is #2000000*{(1+0.02/12)^12-1}~=40368.71# dollars.

So, what happens if the account compounds continuously?
The interest after a year would be
#lim_(n->oo) 2000000*{(1+0.02/n)^n-1}#.

You know the formula
#lim_(n->oo) (1+1/n)^n = e#.

Then,
#lim_(n->oo) (1+0.02/n)^n#
#=lim_(t->oo) (1+1/t)^(0.02t)# (let #t=50n#)
#=e^0.02#.

Therefore, the interest is #2000000*(e^0.02-1)~=40402.68# dollars.