3.8 CHAPTER 3. SEQUENCES AND SERIES
k = 1 : 23 − 13 = 3(1)^2 + 3(1) + 1
k = 2 : 33 − 23 = 3(2)^2 + 3(2) + 1
k = 3 : 43 − 33 = 3(3)^2 + 3(3) + 1
..
.
k = n : (n + 1)^3 −n^3 = 3n^2 + 3n + 1
If we add all the terms on the right and left, we arrive at
(n + 1)^3 − 1 =
�n
i=1
(3i^2 + 3i + 1)
n^3 + 3n^2 + 3n + 1− 1 = 3
�n
i=1
i^2 + 3
�n
i=1
i +
�n
i=1
1
n^3 + 3n^2 + 3n = 3
�n
i=1
i^2 +
3 n
2
(n + 1) +n
�n
i=1
i^2 =
1
3
[n^3 + 3n^2 + 3n−
3 n
2
(n + 1)−n]
=
1
3
(n^3 + 3n^2 + 3n−
3
2
n^2 −
3
2
n−n)
=
1
3
(n^3 +
3
2
n^2 +
1
2
n)
=
n
6
(2n^2 + 3n + 1)
Therefore,
�n
i=1
i^2 =
n
6
(2n + 1)(n + 1)
3.8 Finite Geometric Series EMCAA
When we sum a knownnumber of terms in a geometric sequence, we get a finite geometric series.
We can write out each term of a geometric sequence in the general form:
an= a 1 .rn−^1 (3.27)
where
- n is the index of the sequence;
- anis the nth-term of the sequence;
- a 1 is the first term;
- r is the common ratio (the ratio of any term to theprevious term).
