taylor series 2 lecture participation: problem 1\n(2 points)\nthe function ( f(x)=ln (1 + 5x) ) has a…

taylor series 2 lecture participation: problem 1\n(2 points)\nthe function ( f(x)=ln (1 + 5x) ) has a maclaurin series. find the first 4 nonzero terms in the series, that is write down the taylor polynomial with 4 nonzero terms.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\npage generated october 21, 2025, 10:40:14 pm edt\nwebwork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20\nthe webwork project

taylor series 2 lecture participation: problem 1\n(2 points)\nthe function ( f(x)=ln (1 + 5x) ) has a maclaurin series. find the first 4 nonzero terms in the series, that is write down the taylor polynomial with 4 nonzero terms.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\npage generated october 21, 2025, 10:40:14 pm edt\nwebwork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20\nthe webwork project

Answer

Answer:

$5x-\frac{25x^{2}}{2}+\frac{125x^{3}}{3}-\frac{625x^{4}}{4}$

Explanation:

Step1: Recall the Maclaurin series of $\ln(1 + t)$

The Maclaurin series of $\ln(1 + t)=\sum_{n = 1}^{\infty}\frac{(- 1)^{n+1}}{n}t^{n}=t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\frac{t^{4}}{4}+\cdots$, for $|t|\lt1$.

Step2: Substitute $t = 5x$

Substitute $t = 5x$ into the series. When $n = 1$: $\frac{(-1)^{1 + 1}}{1}(5x)^{1}=5x$. When $n = 2$: $\frac{(-1)^{2+1}}{2}(5x)^{2}=-\frac{25x^{2}}{2}$. When $n = 3$: $\frac{(-1)^{3 + 1}}{3}(5x)^{3}=\frac{125x^{3}}{3}$. When $n = 4$: $\frac{(-1)^{4+1}}{4}(5x)^{4}=-\frac{625x^{4}}{4}$.