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3 years ago

The first three terms of an arithmetic sequence are as follows.

Mathematics

1 answer:

Montano1993 [528]3 years ago

4 0

Answer:

The next two terms and 5 and 8.

Step-by-step explanation:

Find the recursive formula (tₙ = tₙ₋₁ + d) of -4, -1, 2.

Arithmetic sequences have a common difference, "d" , between terms.

To find the common difference, subtract a term value from the term value after it.

-1 - -4 = 3

2 - -1 = 3

The common difference is 3.

To find the next term, add 3 to the previous term.

The recursive formula is tₙ = tₙ₋₁ + 3

tₙ = tₙ₋₁ + 3

t₄ = t₃ + 3

t₄ = 2 + 3

t₄ = 5

tₙ = tₙ₋₁ + 3

t₅ = t₄ + 3

t₅ = 5 + 3

t₅ = 8

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.