Question

coefficient of x^5 in mclaureins expansion of ln(1+x) is​

Answer 1

Answer: sorry wrong que

Step-by-step explanation:

Answer 2

Answer:

Coeffiecient of $x^5$ in the maclaurin expansion of ln(1+x)  = $\frac{1}{5}$.

Step-by-step explanation:

Maclaurin expansion of a function is the polynomial series expansion of the function with a reference point as zero till infinite terms with the coefficients in the terms of derivative of function about the reference point.

The maclaurins expansion of ln(1+x) is given as

                      $ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-....$

From the above expansion we get to know that the coeffiecient of $x^5$ in the maclaurin expansion of ln(1+x) is  $\frac{1}{5}$.