A a ¹2-5, find £r(:) = En n=l n=1 b Given that Σ

Mathematics

1 answer:

VikaD [51]2 years ago

6 0

$\sum^{\infty}_{n=1} (a/b)^n=5 \\ \\ =\frac{a/b}{1-\frac{a}{b}}=5 \\ \\ \frac{a}{b-a} =5 \\ \\ \frac{a}{b}=\frac{5}{6}$

So, we need to find

$\sum^{\infty}_{n=1} n(5/6)^n$

Let this sum be S.

Then,

$S=(5/6)+2(5/6)^2 +3(5/6)^3+\cdots \\ \\ \frac{5}{6}S=(5/6)^2 + 2(5/6)^3+\cdots \\ \\ \implies \frac{1}{6}S=(5/6)+(5/6)^2+(5/6)^3+\cdots=5 \\ \\ \implies S=\boxed{30}$

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For the composite function, identify an inside function and an outside function and write the derivative with respect to x of th

alexira [117]

Answer:

The inner function is $h(x)=4x^2 + 8$ and the outer function is $g(x)=3x^5$ .

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

Step-by-step explanation:

A composite function can be written as $g(h(x))$ , where $h$ and $g$ are basic functions.

For the function $f(x)=3(4x^2+8)^5$ .

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have $4x^2+8$ inside parentheses. So $h(x)=4x^2 + 8$ is the inner function and the outer function is $g(x)=3x^5$ .

The chain rule says:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

It tells us how to differentiate composite functions.

The function $f(x)=3(4x^2+8)^5$ is the composition, $g(h(x))$ , of

outside function: $g(x)=3x^5$

inside function: $h(x)=4x^2 + 8$

The derivative of this is computed as

$\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4$