prove that the square of every term is a term of arithematic sequence 5,9,13...
Answers
Step-by-step explanation:
let there be any odd number 2k+1
squaring
4(k^2+k) +1
since the sequence includes all terms of the form 4x+1 for any +ve number x.
the sequence contains all squares.
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Answer:
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Step-by-step explanation:
The Sigma Notation
The Greek capital sigma, written S, is usually used to represent the sum of a sequence. This is best explained using an example:
This means replace the r in the expression by 1 and write down what you get. Then replace r by 2 and write down what you get. Keep doing this until you get to 4, since this is the number above the S. Now add up all of the term that you have written down.
This sum is therefore equal to 3×1 + 3×2 + 3×3 + 3×4 = 3 + 6 + 9 + 12 = 30.
3
S 3r + 2
r = 1
This is equal to:
(3×1 + 2) + (3×2 + 2) + (3×3 + 2) = 24 .
The General Case
n
S Ur
r = 1
This is the general case. For the sequence Ur, this means the sum of the terms obtained by substituting in 1, 2, 3,... up to and including n in turn for r in Ur. In the above example, Ur = 3r + 2 and n = 3.
Arithmetic Progressions
An arithmetic progression is a sequence where each term is a certain number larger than the previous term. The terms in the sequence are said to increase by a common difference, d.
For example:
3, 5, 7, 9, 11, is an arithmetic progression where d = 2. The nth term of this sequence is 2n + 1 .
In general, the nth term of an arithmetic progression, with first term a and common difference d, is: a + (n - 1)d . So for the sequence 3, 5, 7, 9, ... Un = 3 + 2(n - 1) = 2n + 1, which we already knew.
The sum to n terms of an arithmetic progression
This is given by:
Sn = ½ n [ 2a + (n - 1)d ]
You may need to be able to prove this formula. It is derived as follows:
The sum to n terms is given by:
Sn = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) (1)
If we write this out backwards, we get:
Sn = (a + (n – 1)d) + (a + (n – 2)d) + … + a (2)
Now let’s add (1) and (2):
2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + … + [2a + (n – 1)d]
So Sn = ½ n [2a + (n – 1)d]
Example
Sum the first 20 terms of the sequence: 1, 3, 5, 7, 9, ... (i.e. the first 20 odd numbers).
S20 = ½ (20) [ 2 × 1 + (20 - 1)×2 ]
= 10[ 2 + 19 × 2]
= 10[ 40 ]
= 400
Geometric Progressions
A geometric progression is a sequence where each term is r times larger than the previous term. r is known as the common ratio of the sequence. The nth term of a geometric progression, where a is the first term and r is the common ratio, is:
arn-1
For example,
in the following geometric progression, the first term is 1, and the common ratio is 2:
1, 2, 4, 8, 16, ...
The nth term is therefore 2n-1
The sum of a geometric progression
The sum of the first n terms of a geometric progression is:
a(1 - rn )
1 – r
We can prove this as follows:
Sn = a + ar + ar2 + … + arn-1 (1)
Multiplying by r:
rSn = ar + ar2 + … + arn (2)
(1) – (2) gives us:
Sn(1 – r) = a – arn (since all the other terms cancel)
And so we get the formula above if we divide through by 1 – r .
Example
What is the sum of the first 5 terms of the following geometric progression: 2, 4, 8, 16, 32 ?
S5 = 2( 1 - 25)
1 - 2
= 2( 1 - 32)
-1
= 62