PnC/binomial, sum of choice function series
I don't really know where to start
3 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Hint: Start of by writing C0-C1+C2...=0
1. Write the above down and write C11=C9, C12=C8...C20=C0 and so on
2. Now what you have is 2(C0-C1+C2-...+C10)-C10=0. So 2S=C10