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Novosadov [1.4K]
3 years ago
10

Stoplight a changes color every 30 seconds and stoplight b changes every 40 seconds. after how many seconds will the stoplights

change colors at the same time again?
Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

After 120 seconds

Step-by-step explanation:

Here, we want to know the number of seconds after which both spotlights will change colors at the same time again.

Basically, what we want to do here is to

calculate the value with which 30 and 40 have the same multiples or we want to

calculate the lowest common multiple of 30 and 40

The multiples of 30 are 30,60,90,120 etc

The multiples of 40 are 40,80,120 etc

We can see that the lowest common multiple of both is 120

So therefore the spotlights will change colors at the same time after 120 seconds

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I guess the series is

\displaystyle\sum_{n=1}^\infty\frac{2^nn!}{n^n}

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\right|=2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n

Recall that

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In our limit, we have

\dfrac n{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac1{n+1}

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\implies\displaystyle2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=2\frac{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}}{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)}=\frac{2e}1=2e

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On the chance that you meant to write

\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}

we have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2

=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0

which is less than 1, so this series is absolutely convergent.

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