IPL Auction

I have 2 way to solve this problem. First, just say that, $x$ is the amount of money before the bidding. we can get $x$ itself by expressed :

$x$ = $\frac{x}{2} + \frac{1}{2} +$ amount of money after bidding.

By that formula, we can get the amount of money before the bidder bid for the 11th player

$\frac{2x}{2}$ = $\frac{x}{2}$ + $\frac{1}{2}$ + 0 ................... $\rightarrow$ ( we add 0 because after bidding the 11th player, they remain nothing)

$2x$ = $x + 1$

$x$ = 1

Means that before bidding for the 11th player ( or we can say, after the bid for the 10th player ) the amount of money is ₹1 billion.

Do the equation above for 11 times, you can get the amount of money before the 1st bide, $2047$ billions.

Or maybe you are too lazy to do that, you can realise there is a pattern for each amount of money before the bidding. Before the bidding for 11th player, the bidder had ₹1 billion, or $2^1 - 1$ . Before the bidding for the 10th player, the bidder had ₹3 billion, or $2^2 - 1$ . Before the bidding for the 9th player, the bidder had ₹7 billion, or $2^3 - 1$ . So, before the bidding for the 1st player, the bidder had ₹ $2^{11} - 1= 2047$ billions.

The solution is solved recursively from last player to first player; let us take the instance of last player bid, prior to bid , team will pay certain amount + 0.5 billion; and then it is left with no money, i.e. the bid for last player is 1 bn, similarly for 10th player, the player will have now 1.5 bn + other half; and post those two bids the sum is zero and thus the amount is 3 bn; similarly 3.5+ other half of 3.5 ; and so on and so forth; the series represent the following geometric series i.e. 1,2,4,8 ,... (and the total money required is summation of series 1+2+4+8+....+ 1024 i.e. 2047 billions