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.