Question

Question #42814

Answer 1

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.