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lord [1]
3 years ago
15

Use the formula to evaluate the series 6+2+2/3+2/9

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
5 0
If you're talking about adding just these four terms, notice that

S_4=6+2+\dfrac23+\dfrac29
S_4=6\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}\right)
\dfrac13S_4=6\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}\right)

\implies S_4-\dfrac13S_4=6\left(1-\dfrac1{3^4}\right) [tex]\dfrac23S_4=6\left(1-\dfrac1{81}\right)
S_4=9\times\dfrac{80}{81}
S_4=\dfrac{80}9

If you were referring to an infinite sum, so that the pattern continues, note that the same procedure as above can be applied to the nth partial sum:

S_n=6\left(1+\dfrac13+\dfrac1{3^2}+\cdots+\dfrac1{3^n}\right)
\dfrac23S_n=6\left(1-\dfrac1{3^{n+1}}\right)
S_n=9\left(1-\dfrac1{3^{n+1}}\right)

Then as n\to\infty, the exponential term approaches 0, leaving you with

6+2+\dfrac23+\dfrac29+\cdots=9
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