Question

For fun question

Answer 1

We have an arithmetic progression:
an=number of item at row n

an=a₁+(n-1)d

d=common difference=an-a(n-1)=a₂-a₁=2-1=1
n=number of row

In this case:

an=1+(n-1)*1=n

The sum of an arithmetic serie is:
Sn=(a₁+an)n / 2

In this case:
a₁=1  (number of itms in the first row)
an=n  (we have to calculate this before)

Sn=(1+n)n /2=(n+n²)/2

Therefore:
f(n)=Sn=number of items when we have n number of rows

f(n)=(n+n²)/2

Answer: f(n)=(n+n²)/2

To chek:

f(1)=(1+1²)/2=1
f(2)=(2+2²)/2=6/2=3
f(3)=(3+3²)/2=(3+9)/2=12/2=6
....