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#### Series

If the terms of a sequence are connected with each other by a plus or minus sign, then the resulting expression is called a series. Thus, a series is the well-ordered sum of a sequence. The expression 2+4+6+......+12 is the series associated with the sequence 2,4,6.......,12.

The expression 5+10+15+20 is a finite series associated with finite sequence 5, 10, 15, 20 and 2+4+6+8+..... is an infinite series associated with the infinite sequence 2,4,6,8.........

#### Progression

A sequence is called a progression if the functional relation between its any two successive terms is constant.
In the sequence 2,4,6,8...., the difference between any two successive terms is constant. So it an example of progression.
In the series 5+10+20+40+... the ratio of two successive terms is constant So, it is another example of progression.

There are many types of progression. They are

1. Arithmetic progression (A.P)
2. Geometric progression (G.P)
3. Harmonic Progression (H.P)
4. Logarithmic Progression (L.P)
5. Exponential Progression(E.P)

#### Summation notation or Sigma Notation

If a1,a2,a3,....an be a finite sequence, then the series associated with this sequence is a1+a2+a3+....an. The sum of the series is denoted by Sn= a1+ a2+ a3+ . . . . . . . . . an = $${{\LARGE\sum}_{n=1}^n}a_n$$

Here, ∑ is the summation sign. $${{\LARGE\sum}_{n=1}^n}a_n$$ is read as summation of an's when n runs from 1 to n:

Partial sum

Let a1, a2, a3.....an be the n terms of a sequence. Then , the sum of the terms of this sequence is Sn=a1+a2+a3+...+an =$$\sum_{n=1}^n a_n$$Then, the sum S3=a1+a2+a3= $$\sum_{n=1}^n a_n$$is said to be the partial sum of$$\sum_{n=1}^n a_n$$
Lets Sn=a1+a2+a3+.....an be the sum of n terms of a sequence. Then S4=a1+a2+a3+a4and S3= a1+a2+a3 are the 4th and 3rd partial sum of Sn respectively.
Now,
S4- S3= (a1+a2+a3+a4) -( a1+a2+a3) = a4
Hence, an=Sn= Sn-1

• A sequence is called a progression if the functional relation between its any two successive terms is constant.
• If the terms of a sequence are connected with each other by a plus or minus sign, then the resulting expression is called a series. Thus, a series is the well-ordered sum of a sequence.
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#### Click on the questions below to reveal the answers

Solution

Here,
$$\underset{n=1}{\overset{4}{\LARGE\mathrm (\sum)}} (2n + 1)$$ = (2× 1 + 1) + (2× 2 + 1) + (2× 3 + 1) + (2× 4 + 1)
= (2 + 1) + (4 + 1) + (6 + 1) + (8 + 1)
= 3 + 5 + 7 + 9
= 24

Solution:

Here, the series is 2 + 4 + 6 + 8 + 10
Taking the numbers in sequence
t1 t2 t3 t4 t5
The sequence has first common difference
So, nth term is given by 2n.
∴ 2 + 4 + 6 + 8 + 10 =$$\underset{n=1}{\overset{5}{\LARGE\mathrm (\sum)}}2n$$

Solution:

Here,
the series is-3 + 4 - 5 + 6 - 7 + 8
Taking the numbers in sequence,
∴ 3 4 5 6 7 8
Now, nth term will be given by an + b where,
a = first common differnce = 1
b = first term - first difference = 3 - 1 = 2
Thenn the nth term tn = an + b = 1n + 2 = n + 2
Since the sign are - and + alternatively, its nth sign is (-1)n as the first term is minus (-)
So, the nth term = (-1)n ( n + 1)

a= $$\underset{n=1}{\overset{6}{\LARGE\mathrm \sum}}(-1)^n ( n + 1)$$

Solution:

Given,
Series is 3 + 9 + 27 + 81 + 243
Here,
or, a1 = 3 = 31
or, a2 = 9 = 32
or, a= 27 = 33
or, a4 = 81 = 34
or, a5 = 243 = 35
∴ nth term (an) = 3n
Hence,
3 + 9 + 27 + 81 + 243 = $$\underset{n=1}{\overset{5}{\LARGE\mathrm (\sum)}}$$3n

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• ### Rewrite the 3 + 6 + 9 + 12 using  sigma ((sum)) notation.

(underset{n=5}{overset{3}{LARGEmathrm (sum)}})2n
(underset{n=2}{overset{4}{LARGEmathrm (sum)}})3n
(underset{n=1}{overset{4}{LARGEmathrm (sum)}})3n
(underset{n=2}{overset{5}{LARGEmathrm (sum)}})
• ### Rewrite the 1 + 4 + 7 + 10 + 13 + 16 using sigma ((sum)) notation.

(underset{n=1}{overset{6}{LARGEmathrm (sum)}})3n - 2
(underset{n=2}{overset{5}{LARGEmathrm (sum)}})2n - 3
(underset{n=1}{overset{5}{LARGEmathrm (sum)}})4n - 1
(underset{n=3}{overset{6}{LARGEmathrm (sum)}})3n - 1
• ### Rewrite the 25 + 20 + 15 + 10 + 5 using sigma ((sum)) notation.

(underset{n=1}{overset{5}{LARGEmathrm (sum)}})30 - 5n
(underset{n=2}{overset{5}{LARGEmathrm (sum)}})25 - n
(underset{n=5}{overset{6}{LARGEmathrm (sum)}})n - 20
(underset{n=2}{overset{5}{LARGEmathrm (sum)}})
• ### Rewrite the 8 + 4 + 2 + 1 + (frac{1}{2}) + (frac{1}{4}) + (frac{1}{8})

(underset{n=2}{overset{5}{LARGEmathrm (sum)}})7n
(underset{n=2}{overset{5}{LARGEmathrm (sum)}}) 7n2
(underset{n=4}{overset{7}{LARGEmathrm (sum)}})3 - n2
(underset{n=1}{overset{7}{LARGEmathrm (sum)}})8( (frac{1}{2}))n-1
• ### Rewrite the (frac{3}{5}) + (frac{4}{6}) + (frac{5}{7}) + (frac{6}{8}) + (frac{7}{9})

(underset{n=1}{overset{5}{LARGEmathrm (sum)}}) (frac{2 + n}{5})
(underset{n=5}{overset{n}{LARGEmathrm (sum)}}) 5 + n
(underset{n=1}{overset{7}{LARGEmathrm (sum)}}) (frac{n + 2}{n + 4})
(underset{n=2}{overset{5}{LARGEmathrm (sum)}}) (frac{2}{1 + n})
• ### Rewrite the a + ar + ar2 + ar3 + ar4  using sigma ((sum)) notation.

(underset{n=1}{overset{6}{LARGEmathrm (sum)}}) arn - 1
(underset{n=6}{overset{2}{LARGEmathrm (sum)}}) arn - 3
(underset{n=2}{overset{6}{LARGEmathrm (sum)}}) arn - 6
(underset{n=1}{overset{3}{LARGEmathrm (sum)}}) arn - 2
• ### Rewrite the - 1 + 2 - 3 + 4 - 5 + 6 using sigma (sum) notation.

(underset{n=2}{overset{5}{LARGEmathrm (sum)}}) 2n
(underset{n=1}{overset{6}{LARGEmathrm (sum)}}) (-1). n
(underset{n=1}{overset{2}{LARGEmathrm (sum)}}) (-4)n
(underset{n=3}{overset{4}{LARGEmathrm (sum)}}) 5n - 1
• ### Rewrite the (x-1) + (x-2)2 + (x-3)3 + . . . . . . .  + (x-10)10

(underset{n=3}{overset{5}{LARGEmathrm (sum)}}) (x + n)
(underset{n=1}{overset{4}{LARGEmathrm (sum)}}) (x - n)2
(underset{n=1}{overset{10}{LARGEmathrm (sum)}}) (x - n)n
(underset{n=2}{overset{4}{LARGEmathrm (sum)}}) (x + n)10

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