1. ## Urgent! AP/GP

Hey I need a solution for part (b) only:

A bank has an account for investors. Interest is added to the account at the end of each year at a fixed rate of 5% of the amount in the account at the beginning of that year. A man and a woman both invest money.

(b) The woman decides that, to assist her in her everyday expenses, she will withdraw interest as soon as it has been added. She invests $y at the beginning of each year. Show that, at the end of n years, she will have received a total of$[n(n+1)y]/40 in interest.

Thanks if you could help, I'm having my math paper in about 8 hours' time!

2. Hello, Margaritas!

At a bank interest is added to the account at the end of each year
at a fixed rate of 5% of the amount in the account at the beginning of that year.

(b) A woman decides that she will withdraw interest as soon as it has been added.
She invests $y at the beginning of each year. Show that, at the end of n years, she will have received a total of$[n(n+1)y]/40 in interest.

At the end of year-1, she gets 5% of $y = 0.05y = y/20 dollars in interest. She invests$y more; at the end of year-2, she gets: 0.05(2y) = 2y/20 dollars in interest.

She invests \$y more; at the end of year-3, she gets: 0.05(3y) = 3y/20 dollars in interest.

. . . and so on.

By the end of the nth year, she gets a total of:
. . y/20 + 2y/20 + 3y/20 + . . . ny/20 dollars in interest.

. . So she has: .T .= .y(1 + 2 + 3 + ... + n)/20 dollars.

The sum of the first n integers is: .1 + 2 + 3 + ... + n .= .n(n+1)/2

. . Therefore: .T .= .y[n(n+1)/2]/20 .= .n(n+1)y/40 dollars.

3. Thanks Soroban, I get it now!