Arithmetico-Geometric Series &
Some Special Sequences
A series each term of which is formed by multiplying the corresponding
terms of an A.P. and G.P. is called arithmetico-geometric series.
The general or standard form of such a series is a, (a +d) r, (a +2 d) r², ...
The sum of first n natural numbers
Let Sn (or 
n²) denote the sum of first n
natural numbers, then
Sn = 1 +2 +3 +... +n
This is an A.P. with a = 1, d = 1 and n terms.
Using formula Sn = n(a +l)/2,
Sn = n(1 +n)/2 = n(n +1)/2 or
n= n(n +1)/2
The sum of squares of first n natural numbers
Let Sn (or n²) denote the sum of
squares of the first n natural numbers, then
Sn = 1² +2² +3² +... +n².
Consider the identity x³ -(x -1)³ = 3 x² -3 x +1 ...(i)
Putting x = 1, 2, 3, ..., n in (i), we get
1³ -0³ = 3. 1² -3. 1 +1
2³ -1³ = 3. 2² -3. 2 +1
3³ -2³ = 3. 3² -3. 3 +1
..........................................
..........................................
n³ -(n -1)³ = 3. n² -3. n +1
Adding column wise, we get
n³ -0³ = 3(1² +2² +... +n²) -3 (1 +2 +... +n) + n
n³ = 3 n² -3 n +n
therefore n² = n(n +1)(2n +1)/6
The sum of cubes of the first n natural numbers
Let Sn or n³ denote the sum of the
cubes of the first n natural numbers.
Then Sn = 1³ +2³ +3³ +... +n³.
Consider the identity x4 -(x -1)4 = 4 x³ -6 x² +4 x -1,
we get
Method of differences
In some series, the differences of successive terms (Tn and Tn -1) may be in some particular sequence -A.P., G.P., A.G.P. etc. In these cases, the sum of series can be found by a technique known as "method of differences".
Illustrative Examples
Example
Sum to n terms the series
2 +5 x +8 x² +11 x³ +..., |x| < 1. Deduce the sum to infinity.
Solution
The given series is formed by multiplying corresponding terms of A.P. 2, 5,
8, 11, ... and G.P. 1, x, x², x³, ...
Hence n th term of given arithmetico-geometric series is
[2 +(n -1) 3]. xn -1 = (3 n -1) xn -1
Let Sn = 2 +5 x +8 x² +11 x³ +... +(3 n -4) xn -2 +(3 n -1) xn -1
x Sn = 2 x +5 x² +8 x³
+...
+(3 n -4) xn -1 +(3 n -1) xn
(1 -x) Sn = 2 +(3 x +3 x² +3 x³ +... +3 xn -1) -(3n -1) xn
= 2 +3 x (1 +xn -1)/(1-x)
-(3 n -1) xn
Sn = 2/(1-x) + 3 x (1 +xn -1)/(1-x)² -(3 n -1)
xn/(1-x)
To find limiting sum S, we note that
xn -1, xn => 0, n xn => 
because |x| < 1
Therefore S = 2/(1-x) + 3x/(1 -x)² = (2 + x)/(1 -x)²
Example
Sum to infinity the series 1² +2². x + 3². x² +..., |x| < 1.
Solution
Though the given series is not an arithmetico-geometric series, it can be
summed up using a similar method.
Let S = 1 +4 x +9 x² +16 x³ +... x
Sx = x +4 x² + 9 x³ +...
=> (1 -x) S = 1 +3 x +5 x² +17 x³ +...,
which is an arithmetico-geometric series
x (1 -x) S = x +3 x²+ 5 x³ +...
=> (1 -x)² S = 1 +2 x +2 x² +2 x³ +...
= 1 + 2x/(1-x) = (1 +x)/(1-x).
=> S = (1 +x)/(1 -x)³
Example
Find the sum of the series 2. 5 +5. 8 +8. 11 +... to n terms.
Solution
This series is formed by multiplying the corresponding terms of sequences
2, 5, 8, ... and 5, 8, 11, ... both of which are A.P.s.
Now nth term of first A.P. = 2 +(n -1)3 = 3n -1
and nth term of second A.P. = 5 +(n -1)3 = 3n +2
Hence nth term of the given series,
Tn = (3 n -1)(3 n +2) = 9 n² +3 n -2
Thus the required sum,
Sn = Tn = (9
n² +3 n -2) = 9 n² +3 n -2 n
= 9 . n (n +1)(2n +1)/6 +3n (n +1)/2 -2 n
= n (6 n² +12 n +2)/2 = n (3 n² +6 n +1)/2
Example
Find the sum of first n terms of the series 1 +(1 +2) +(1 +2 +3) +.........
Solution
nth term of the given series,
Tn = 1 +2 +3 +... up to n = n (n +1)/2 = n²/2 +n/2
Sum of first n terms of the given series
=[ n (n +1)/12](2n +1 +3) = n(n +1)(n +2)/6
Example
Find the nth term of series 1 +3 +7 +13 +..., and hence find the sum of first n terms.
Solution
Here the differences of successive terms are 2, 4, 6, ... which form an A.P.
Let Tn denote the nth term and Sn denote the sum
of first n terms.
Then Sn = 1 +3 +7 +13 +... +Tn -1 +Tn
Also Sn = 1 +1 +3 +17 +............ +Tn -1 +Tn
0 = 1 +2 +4 +6
+... upto n terms -Tn
Tn = 1 +(2 +4 +6 +... upto n-1 terms)
= 1 + (n -1)[2. 2 +(n -2)2]/2 = n² -n +1
Sn = n²
- n +n
=
n (n +1)(2 n +1)/6 n (n +1)/2 +n
= n [2 n² +3 n +1 -3 (n +1) +6]/6
= n (2n² +4)/6 = n (n²+2)/3
Example
Find the n th term, sum of n terms, and sum to infinity of the series
1/2.5 +1/5.8 +1/8.11+ ...
Solution
Tn = 1/ [(n th terms of 2, 5, 8,....)(n th term of 5, 8, 11,...)]
= 1/[(2 +(n -1) 3)(5 +(n -1) 3)]
= 1/ [(3n -1)(3 n +2)]
=
Putting n = 1, 2, 3, ..., n, we get
T1 =
T2 =
.....................
Tn =
Adding, Sn =
Also sum to infinity =
= 
Exercise
- Sum to infinity the following series:
(i) 1 +3/2 +5/2² + 7/2³ +
(ii) 3 +5/4 + 7/4² + ....
(iii) 1 +4 x² +7 x4 + 10 x6 +..., |x| < 1 - Sum to n terms the following series:
(i) 2. 1 +3. 2 +4. 4 +5. 8 +...
(ii) 1. 2 +2. 2² +3. 2³ +4. 24 +... - Sum to n terms the following series:
(i) 1 +(1 +3) +(1 +3 +5) +...
(ii) 1 + (1 +2)/2 +(1 +2 +3)/3 +...
(iii) 2² +5² +8² +...
(iv) 1. 3 +3. 5 +5. 7 +... - (i) Find the sum of series 2² +5² +8² +... upto 15 terms.
(ii) Find the sum of squares of first 100 odd natural numbers.
(iii) Find the sum of cubes of first 100 odd natural numbers.
(iv) Find the sum of the products of first n natural numbers taken two at a time. - Sum to n terms the following series:
(i) 1 +5 +14 +30 +55 +...
(ii) 3 +7 +14 +24 +37 +...
(iii) 6 +9 +16 +27 +42 +...
(iv)
(v)
(vi) 1³ +(1³+2³)/2 +(1³ +2³ +3³)/3 +...
(vii) 4 +44 +444 +...
(viii) 0·5 +0·55 +0·555 +... - Sum the following series to n terms 5 +7 +13 +31 +85 +...
- Prove that sum of cubes of any number of consecutive natural numbers is
always divisible by the sum of these numbers.
[Hint. Let the m consecutive natural numbers be n +1, n +2, ..., n +m]
Answers
1. (i) 6 (ii) 44/9 (iii) 1/(1 -x²) +3x²/(1 -x²)²2. (i) n. 2n (ii) 2 +(n -1) 2 n +1
3. (i) n (n +1)(2n +1)/6 (ii) n (n +3)/4
(iii) n (6 n² +3 n -1)/2 (iv) n (4 n² +6 n -1)/3
4. (i) 10455 (ii) 1333300 (iii) 199990000
(iv) (n -1)n(n +1)(3n +2)/24 [Hint. Consider (1 +2 +... + n)²]
5. (i) n (n +1)² (n +2)/12 (ii) n(n² +n +4)/2
(iii) n(4 n² -3 n +35)/6 (iv) 2 n -2 + 1/2n-1
(v)
(vi) n (n +1)(n +2)(3n +5)/48
(vii) 4(10n +1 -9n -10)/81 (viii) 5(9n -1 +1/10n)/81
6. (3n +8 n -1)/2