Please scroll down to get to the study materials.
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.........
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
If a_{1},a_{2},a_{3},....a_{n} be a finite sequence, then the series associated with this sequence is a_{1}+a_{2}+a_{3}+....a_{n}. The sum of the series is denoted by Sn= a_{1}+ a_{2}+ a_{3}+ . . . . . . . . . a_{n} = \({{\LARGE\sum}_{n=1}^n}a_n\)
Here, ∑ is the summation sign. \({{\LARGE\sum}_{n=1}^n}a_n\) is read as summation of a_{n}'s when n runs from 1 to n:
Partial sum
Let a_{1}, a_{2}, a_{3}.....a_{n} be the n terms of a sequence. Then , the sum of the terms of this sequence is S_{n}=a_{1}+a_{2}+a_{3}+...+a_{n} =\(\sum_{n=1}^n a_n\)Then, the sum S_{3}=a_{1}+a_{2}+a_{3}= \(\sum_{n=1}^n a_n\)is said to be the partial sum of\(\sum_{n=1}^n a_n\)
Lets S_{n}=a_{1}+a_{2}+a_{3}+.....a_{n} be the sum of n terms of a sequence. Then S_{4}=a_{1}+a_{2}+a_{3}+a_{4}and S_{3}= a_{1}+a_{2}+a_{3} are the 4^{th} and 3^{rd} partial sum of S_{n} respectively.
Now,
S_{4}- S_{3}= (a_{1}+a_{2}+a_{3}+a_{4}) -( a_{1}+a_{2}+a_{3}) = a_{4}
Hence, a_{n}=S_{n}= S_{n-1}
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
t_{1} t_{2} t_{3} t_{4} t_{5}
The sequence has first common difference
So, n^{th} 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, n^{th} term will be given by an + b where,
a = first common differnce = 1
b = first term - first difference = 3 - 1 = 2
Thenn the n^{th} term t_{n} = an + b = 1n + 2 = n + 2
Since the sign are - and + alternatively, its n^{th} sign is (-1)^{n} as the first term is minus (-)
So, the n^{th} term = (-1)^{n} ( n + 1)
a_{n }= \(\underset{n=1}{\overset{6}{\LARGE\mathrm \sum}}(-1)^n ( n + 1)\)
Solution:
Given,
Series is 3 + 9 + 27 + 81 + 243
Here,
or, a_{1} = 3 = 3^{1}
or, a_{2} = 9 = 3^{2}
or, a_{3 }= 27 = 3^{3}
or, a_{4} = 81 = 3^{4}
or, a_{5} = 243 = 3^{5}
∴ n^{th} term (a_{n}) = 3^{n}
Hence,
3 + 9 + 27 + 81 + 243 = \(\underset{n=1}{\overset{5}{\LARGE\mathrm (\sum)}}\)3^{n}
Rewrite the 3 + 6 + 9 + 12 using sigma ((sum)) notation.
Rewrite the 1 + 4 + 7 + 10 + 13 + 16 using sigma ((sum)) notation.
Rewrite the 25 + 20 + 15 + 10 + 5 using sigma ((sum)) notation.
Rewrite the 8 + 4 + 2 + 1 + (frac{1}{2}) + (frac{1}{4}) + (frac{1}{8})
Rewrite the (frac{3}{5}) + (frac{4}{6}) + (frac{5}{7}) + (frac{6}{8}) + (frac{7}{9})
Rewrite the a + ar + ar^{2} + ar^{3} + ar^{4} using sigma ((sum)) notation.
Rewrite the - 1 + 2 - 3 + 4 - 5 + 6 using sigma (sum) notation.
Rewrite the (x-1) + (x-2)^{2} + (x-3)^{3} + . . . . . . . + (x-10)^{10}
ASK ANY QUESTION ON Series
No discussion on this note yet. Be first to comment on this note