This is a question from a recent ACJC JC2 test.

A philanthropist started a donation matching programme to encourage more people to donate regularly to a particular charity.

(a) For a person who donates $a in the first month, and for each subsequent month donates$ $b^2$ more than the previous month, the philanthropist will donate $a in the first month and for each subsequent month, b times the amount he donated the previous month. (i) John is a regular donor of the charity. Find the values of a and b such that the philanthropist donates$20 in the third month, and ten times more than John in the seventh month.

(ii) Find the total amount of money donated by John and the philanthropist in one year, leaving your answer to the nearest dollar.

(b) In a revised donation matching programme, if a donor makes a monthly donation of $c, the philanthropist will donate according to the following plan: First month : No donation ($0\%$ donation rate) Second month : $10 \%$ of the total amount donated by the donor in the first two months Third month : $20 \%$ of the total amount donated by the donor in the first three months Fourth month : $30 \%$ of the total amount donated by the donor in the first four months and so on. (i) Show that the total amount of money the philanthropist will donate at the end of the $4^{th}$ month is$2c.

(ii) By expressing the total amount of money, the philanthropist will donate at the end of the $n^{th}$ month as a summation, and using the result $\sum_{r=1}^n r^2 = \frac{n}{6}(n+1)(2n+1)$, show that the total amount of money the philanthropist will donate at the end of the $n^{th}$ month is \$ $\frac{(n-1)n(n+1)}{30} c$.