What is ∑n=15[5(−2)n−1] equal to?

Mathematics

1 answer:

kicyunya [14]2 years ago

6 0

Answer:

Although the notation is sloppy, I will assume based on former questions of the same that "∑n=15[5(−2)^n−1]" translates mathematically to . Please correct me, though, if this in incorrect.

Anyway, there is neat formula that will deal with question very quickly. It is the following:

Sn is the sum of the geometric series

a1 is the first term in the series

r is the common ratio

n is the number of terms in the finite geometric series

Let's use this formula to our advantage! However, there are two variables that we do not know: a1 and n. Let's start with a1.The first term in the series is when n=1, so let's determine what that is.

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Masja [62]

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2 years ago

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Harrizon [31]

Answer: If a number is rational, it can be represented as a simple fraction, it can be written as a terminating decimal, and a repeating decimal.

If a number is irrational, it cannot be represented as a simple fraction, it cannot be written as a terminating decimal, and not as a repeating decimal. It never ends, never repeats, just like the constant pi.

Step-by-step explanation:

So if a number is rational, it can be represented as a simple fraction, it can be written as a terminating decimal, and a repeating decimal.

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3 years ago

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

Answer:

ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

$\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}$