The sum of the converging series with the first term as -4 is -6.
Given to us,
- Series, [tex]\sum^{\infty }_{n=1}=-4(\dfrac{1}{3})^{(n-1)}}[/tex]
where, n=1
To find
a.) The first four terms of the series,
[tex]\sum^{\infty }_{n=1}=-4(\dfrac{1}{3})^{(n-1)}}[/tex]
first term,
n=1,
[tex]=-4(\dfrac{1}{3})^{(1-1)}}\\=-4(\dfrac{1}{3})^{(0)}}\\=-4[/tex]
Second term,
n=2,
[tex]=-4(\dfrac{1}{3})^{(2-1)}}\\=-4(\dfrac{1}{3})^{(1)}}\\=-4\times \dfrac{1}{3}\\=\dfrac{-4}{3}[/tex]
Third term,
n=3,
[tex]=-4(\dfrac{1}{3})^{(3-1)}}\\=-4(\dfrac{1}{3})^{(2)}}\\=-4\times \dfrac{1}{9}\\=\dfrac{-4}{9}[/tex]
Fourth term,
n=4,
[tex]=-4(\dfrac{1}{3})^{(4-1)}}\\=-4(\dfrac{1}{3})^{(3)}}\\=-4\times \dfrac{1}{27}\\=\dfrac{-4}{27}[/tex]
To find
b.) The series diverge or converge,
The series is a converging series as the n value increases the value of the series decreases, because, the more the value of n the smaller number we will get. And, as we can see the n is in the denominator.
Hence, the series is a converging series.
To find
c.) The sum of the series,
We know that sum of a series is given as [tex]\bold{\dfrac{a}{(1 - r)}}[/tex].
where, a is the first term and r is the common ratio.
In this series,
first term, a = - 4
the common ratio, r = [tex]\dfrac{1}{3}[/tex]
Thus, the sum of the series = [tex]{\dfrac{a}{(1 - r)} = {\dfrac{-4}{(1 - {\frac{1}{3})}} = \dfrac{-4}{\frac{2}{3}}= -6[/tex].
Hence, the sum of the converging series with the first term as -4 is -6.
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