List the common factors of the numbers 9 and 18
1 answer:
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By inspection, it's clear that the sequence must converge to 
because
when
is arbitrarily large.
Now, for the limit as
to be equal to is to say that for any 
, there exists some 
such that whenever 
, it follows that
From this inequality, we get
As we're considering, we can omit the first inequality.
We can then see that choosing
will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that 
.
when
Now, for the limit as
From this inequality, we get
As we're considering
We can then see that choosing
The series 7 + 16 + 25 +34 +43 +52 + 61 is an illusration of arithmetic series
The sigma notation of the series is:
The series is given as:
7 + 16 + 25 +34 +43 +52 + 61
The above series is an arithmetic series, with the following parameters
- First term, a = 7
- Common difference, d = 9
- Number of terms, n = 7
Start by calculating the nth term using:
a(n) = a + (n - 1) * d
This gives
a(n) = 7 + (n - 1) * 9
Evaluate the product
a(n) = 7 - 9 + 9n
Evaluate the difference
a(n) = 9n - 2
So, the sigma notation is:
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