vikas askiitian expert
Last Activity: 13 Years ago
let first peson gets minimum marble = 1
second gets minimum 2 marbles , third gets minimum 3 marbles & so on.......
this is an AP series whose first term is 1 & common difference is 1 ...
now total marbles distributed to n-1 people = k
k = [(n-1)/2][2+(n-2)] = n(n-1)/2
now , last person gets all remaining marbles ..
remaining marbles = total - distributed
= n2 - (n(n-1)/2)
= n(n+1)/2
this is the maximum number of marble which any 1 of them can get ...
approve if u like my ans