chapter  3
Polynomials
Pages 14

Proof: We have ∆qk = qk+1 − qk = (q − 1)qk. So according to (1.10) and (1.9) we get (3.2):

qk = ∑n+1

m qkδk =

qn+1

q − 1 − qm

q − 1 = qn+1 − qm

q − 1 = q m 1− qn−m+1

1− q .

According to an anecdote, C. F. Gauss (1777-1855) went to school at the age of seven. One day the teacher, in order to keep the class occupied,

asked the students to add up all the numbers from one to one hundred. Gauss immediately answered 5050. He wrote the numbers in two lines:

1 + 2 + 3 + · · · + 100, 100 + 99 + 98 + · · · + 1.