A bank has an account for investors. Interest is added to the account at the end of each year at a fixed rate of5%of the amount in the account at the beginning of that year. A man and woman both invest money.

(a) The man decides to invest$xat the beginning of one year and then a further$xat the beginning of the second year and each subsequent year. He also decides that he will not draw any money out of the account, but just leave it, and any interest, to build up. Show that at the end of n years, when the interest for the last year has been added, he will have a total of $$\displaystyle 21(1.05^{n}-1)x$ in his account.

what i tried to do is:

$\displaystyle \frac{x(1.05^{n}-1)}{1.05-1}$

$\displaystyle

=20x(1.05^{n}-1)

$

where did i go wrong?? its is 21...

(b) The woman decides that, to assist her everyday expenses, she will withdraw the interest as soon as it has been added. She invest $y at the beginning of each year. Show that, at the end of n years, she will receive a total of $$\displaystyle \frac{1}{40}$$\displaystyle n(n+1)y$ in interest.

i'm starting to get blur....@"@ is this arithmetic progression or geometric progression?