Question 847528
First child gets 4.
Second child gets 7... and so on.

nth child gets 4+(n-1)*3  or  3n+1 chocolates.

sum(1 to 10) of (3n+1)

3*sum(1 to 10) + 10

{{{3 * ((10)(11)/2) + 10}}}  note we used the closed form  {{{(n(n+1))/2}}}

{{{3* 55 + 10 = 175}}}