The sum of the first n terms os the series is given Sn=2n(n+3). Show that the terms of the series forms an arithmetic progression.
2n(n+3) = (4n)/2 (2*8 + ((4n)-1)*4)
So, if m = 4n,
Sm = m/2 (2*8 + (m-1)*4)
Solve please
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This is not the question iask
To show that the terms of the series form an arithmetic progression, we need to prove that the difference between consecutive terms is constant.
Let's find the expression for the nth term of the series. We are given that Sn, the sum of the first n terms, is equal to 2n(n + 3).
To find the nth term, we can use the formula for the sum of an arithmetic series, which is Sn = (n/2)(a + l), where a is the first term and l is the last term.
In this case, the last term can be written as a + (n - 1)d, where d is the common difference of the arithmetic progression.
So, we have 2n(n + 3) = (n/2)(a + a + (n - 1)d).
Simplifying the equation, we get:
4n(n + 3) = n(a + a + (n - 1)d)
Expanding the equation, we have:
4n^2 + 12n = n(2a + (n - 1)d)
Now, let's compare the coefficients of n^2, n, and the constant terms on both sides of the equation.
On the left side, we have 4n^2 + 12n.
On the right side, we have n(2a + (n - 1)d).
Comparing the coefficients, we get:
4 = 2a
12 = (n - 1)d
From the first equation, we find that a = 2.
Substituting this value into the second equation, we get:
12 = (n - 1)d
Dividing both sides by (n - 1), we have:
d = 12/(n - 1)
So, we have found that the common difference (d) is equal to 12/(n - 1).
Therefore, the terms of the series form an arithmetic progression with a common difference of 12/(n - 1).