Simply using exponents .Then find the products .

Mathematics

1 answer:

aksik [14]4 years ago

6 0

Answer:

$\sqrt[7]{b^{15}}$

Step-by-step explanation:

=> $b^2*\sqrt[7]{b}$

=> $b^2*b^{1/7}$

Where bases are same, powers are to be added.

=> $b^{2+(1/7)}$

=> $b^ {14+1/7}$

=> $b^{15/7}$

=> $\sqrt[7]{b^{15}}$

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$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

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The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

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We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

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