Let T

Then T

Now, 1d

Now, 2d

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 = \({S_n}\) = \({}^n{C_1} \times {T_1}\) + \({}^n{C_2} \times \left( {1{d_1}} \right)\) + \({}^n{C_3} \times \left( {2{d_1}} \right)\) + . . .

Find the general term and the sum of

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 = \({}^n{C_1} \times 12\) + \({}^n{C_2} \times \left( {28} \right)\) + \({}^n{C_3} \times \left( {22} \right)\) + \({}^n{C_4} \times \left( 6 \right)\)

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

Find the sum of the series

3 + 5x + 9x

1st order differences = 2, 4, 6, 8, . . .

2nd order differences = 2, 2, 2, 2...

So Scale of the series = (1 - x)

(1 - x)

Therefore, we find S - 3xS + 3x

S = 3 + 5x + 9x

- 3xS = -9x -15x

3x

-x

⇒ S - 3xS + 3x

⇒ S(1 - 3x + 3x

⇒ \(S = \dfrac{{3 - 4x + 3{x^2}}}{{{{\left( {1 - x} \right)}^3}}}\)

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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 + 3x

Here, \(5{x^4} = 2x \times 4{x^3} - {x^2} \times 3{x^2}\)

So scale of series = \(1 - 2x + {x^2}\)

Find the scale of the series: 2 + 5x + 13x

Let the scale of the relation be 1 - px - qx

Therefore, 13 - 5p - 2q = 0; 35 - 13p - 5q = 0

Solving, p = 5, and q = -6

Thus scale of notation = 1 - 5x + 6x

Find the sum of the series, 1 - 7x - x

Let the scale of the relation be 1 - px - qx

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 - 6x

Let S denote the sum of the sereis. Then

S = 1 - 7x - x

- xS = -x + 7x

-6x

(S - xS - 6x

\( \Rightarrow S = \dfrac{{1 - 8x}}{{1 - x - 6{x^2}}}\)

###

Find the nth term of the series 10, 23, 60, 169, 494, . . .

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.3

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

_{1}, T_{2}, T_{3}... are the terms of a series.Then T

_{2}- T_{1}, T_{3}- T_{2}, T_{4}- T_{3}. . . are called the series of the first order of differences. We denote it by 1d_{1}, 1d_{2}, 1d_{3}. . .Now, 1d

_{2}- 1d_{1}, 1d_{3}- 1d_{2}, 1d_{3}- 1d_{2}, . . . is called the series of the second order of differences. This is denoted as 2d_{1}, 2d_{2}, 2d_{3}. . .Now, 2d

_{2}- 2d_{1}, 2d_{3}- 2d_{2}, 2d_{3}- 2d_{2}, . . . is called the series of the third order of differences. This is denoted as 3d_{1}, 3d_{2}, 3d_{3. . . }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

**terms of the series 12, 40, 90, 168, 280, 432, . . .***n***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 = \({}^n{C_1} \times 12\) + \({}^n{C_2} \times \left( {28} \right)\) + \({}^n{C_3} \times \left( {22} \right)\) + \({}^n{C_4} \times \left( 6 \right)\)

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**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 + 9x

^{2}+ 15x^{3}+ 23x^{4}+ 33x^{5}+ . . .**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 + 3x^{2}- x^{3}Therefore, we find S - 3xS + 3x

^{2}S - x^{3}S. 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 + 9x

^{2}+ 15x^{3}+ 23x^{4}+ 33x^{5}+ . . .- 3xS = -9x -15x

^{2}- 27x^{3}- 45x^{4}- 69x^{5 }. . .3x

^{2}S = 9x^{2}+ 15x^{3}+ 27x^{4}+ 45x^{5}+ . . .-x

^{3}S = - 3x^{3}- 5x^{4}- 9x^{5}- . . .⇒ S - 3xS + 3x

^{2}S - x^{3}S = 3 - 4x + 3x^{2}⇒ S(1 - 3x + 3x

^{2}- x^{3}) = 3 - 4x + 3x^{2}⇒ \(S = \dfrac{{3 - 4x + 3{x^2}}}{{{{\left( {1 - x} \right)}^3}}}\)

###

**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 + 3x

^{2}+ 4x^{3}+ 5x^{4 }+ . . .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 + 13x

^{2}+ 35x^{3}+ . . .**Solution:**Let the scale of the relation be 1 - px - qx

^{2}Therefore, 13 - 5p - 2q = 0; 35 - 13p - 5q = 0

Solving, p = 5, and q = -6

Thus scale of notation = 1 - 5x + 6x

^{2}**Example 4:**Find the sum of the series, 1 - 7x - x

^{2}- 43x^{3}- . . .**Solution:**Let the scale of the relation be 1 - px - qx

^{2}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 - 6x

^{2}Let S denote the sum of the sereis. Then

S = 1 - 7x - x

^{2}- 43x^{3}- . . .- xS = -x + 7x

^{2}+ x^{3}+^{ }. . .-6x

^{2}S = - 6x^{2}+ 42x^{3}+ . . .(S - xS - 6x

^{2}S) = 1 - 8x\( \Rightarrow S = \dfrac{{1 - 8x}}{{1 - x - 6{x^2}}}\)

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**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.3

^{n-1}+ bn + cTo 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 }+ n + 3