# Sum of the series by the method of differences

Let T1, T2, T3 ... are the terms of a series.
Then T2 - T1, T3 - T2, T4 - T3 . . . are called the series of the first order of differences. We denote it by 1d1, 1d2, 1d3. . .
Now, 1d2 - 1d1, 1d3 - 1d2, 1d3 - 1d2, . . . is called the series of the second order of differences. This is denoted as 2d1, 2d2, 2d3. . .
Now, 2d2 - 2d1, 2d3 - 2d2, 2d3 - 2d2, . . . is called the series of the third order of differences. This is denoted as 3d1, 3d2, 3d3. . .
General term of the above series =  ${T_n}$ = ${T_1}$ + $\left( {n - 1} \right) \times \left( {1{d_1}} \right)$ + $\dfrac{{(n - 1)(n - 2)}}{{1.2}} \times \left( {2{d_1}} \right)$ + . . .
(or)
${T_n} = {T_1}$ + ${}^{(n - 1)}{C_1} \times \left( {1{d_1}} \right)$ + ${}^{(n - 2)}{C_2} \times \left( {2{d_1}} \right)$ + . . .

Sum of the terms of the above series =  ${S_n}$ = ${}^n{C_1} \times {T_1}$ + ${}^n{C_2} \times \left( {1{d_1}} \right)$ + ${}^n{C_3} \times \left( {2{d_1}} \right)$ + . . .

Example 1:
Find the general term and the sum of n terms of the series 12, 40, 90, 168, 280, 432, . . .
Solution:
The successive orders of differences are
28, 50, 78, 112, 152, . . .
22,  28,   34,   40, . . .
6,     6,     6, . . .
0,     0, . . .
Here third series of differences are equal. So
General term = $12$ + $28(n - 1)$ + $\dfrac{{22(n - 1)(n - 2)}}{{1.2}}$ + $\dfrac{{6(n - 1)(n - 2)(n - 3)}}{{1.2.3}}$ + . . . = ${n^3} + 5{n^2} + 6n$
(There is no need to simplify the above expression.  To find the nth term, substitute n value and the simplify)
Sum of the terms = ${}^n{C_1} \times 12$ + ${}^n{C_2} \times \left( {28} \right)$ + ${}^n{C_3} \times \left( {22} \right)$ + ${}^n{C_4} \times \left( 6 \right)$

### Solving Series questions using scale concept:

If  ${n^{th}}$ series of differences are equal for a series, then scale of the series =  ${\left( {1 - x} \right)^{n + 1}}$

Example 2:
Find the sum of the series
3 + 5x + 9x2 + 15x3 + 23x4 + 33x5 + . . .
Solution:
1st order differences = 2, 4, 6, 8, . . .
2nd order differences = 2, 2, 2, 2...
So Scale of the series = (1 - x)2+1 = (1 - x)3
(1 - x)3 = 1 - 3x + 3x2 - x3
Therefore, we find S - 3xS + 3x2S - x3S.  When you write the terms in the below mentioned format, automatically, sum of the terms in 4th column, 5th column, ...  after the equal sign will become zero.
S =  3 + 5x + 9x2 + 15x3 + 23x4 + 33x5 + . . .
- 3xS =       -9x -15x2 - 27x3 - 45x4 - 69x5 . . .
3x2S =                9x2 + 15x3 + 27x4 + 45x5 + . . .
-x3S =                         - 3x3 - 5x4 - 9x5 - . . .
⇒ S - 3xS + 3x2S - x3S = 3 - 4x + 3x2
⇒ S(1 - 3x + 3x2 - x3) = 3 - 4x + 3x2
⇒ $S = \dfrac{{3 - 4x + 3{x^2}}}{{{{\left( {1 - x} \right)}^3}}}$

### Solving recurring series:

In a series in which from and after a certain term each term is equal to the sum of a fixed number of the preceding terms multiplied respectively by certain constants is called "recurrence series".
For example, 1 + 2x  + 3x2 + 4x3 + 5x+ . . .
Here, $5{x^4} = 2x \times 4{x^3} - {x^2} \times 3{x^2}$
So scale of series = $1 - 2x + {x^2}$

Example 3:
Find the scale of the series: 2 + 5x + 13x2 + 35x3 + . . .
Solution:
Let the scale of the relation be 1 - px  - qx2
Therefore, 13 - 5p - 2q = 0; 35 - 13p - 5q = 0
Solving, p = 5, and q = -6
Thus scale of notation = 1 - 5x + 6x2

Example 4:
Find the sum of the series, 1 - 7x  - x2 - 43x3 - . . .
Solution:
Let the scale of the relation be 1 - px  - qx2
Then, - 1 + 7p - q = 0; and - 43 + p + 7q = 0
Solving above equations, we get p = 1 and q = 6.
Scale of the relation is 1 - x - 6x2
Let S denote the sum of the sereis. Then
S  =  1 - 7x  - x2 - 43x3 - . . .
- xS   =       -x + 7x2 + x3 + . . .
-6x2S =            - 6x2 + 42x3 + . . .
(S - xS - 6x2S) = 1 - 8x
$\Rightarrow S = \dfrac{{1 - 8x}}{{1 - x - 6{x^2}}}$

### Finding general term when ratio of terms after successive differences:

Example 5:
Find the nth term of the series 10, 23, 60, 169, 494, . . .
Solution:
Successive differences are
13,   37,   109,    335, . . .
24,   72,     216, . . .
Here 2nd order differences ratio is equal and equal to 3.
So General term = a.3n-1 + bn + c
To determine the constants, a, b, c make n equal to 1, 2, 3, successively and equal to the terms in the original series.
Then a + b + c = 10; 3a + 2b + c = 23; 9a + 3b + c = 60
Solving, we get a = 6, b = 1, c = 3
So general term = 6.3(n-1) + n + 3 = 2.3+ n + 3