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Mathematics

1 answer:

Cerrena [4.2K]2 years ago

6 0

I you think about it 12a with the 2 on the rope is actually 144 because it’s multiplied by 12 so maybe that helps?

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What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$