The coefficient of x^3 in the expansion of 3^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $3^x = e^{\log(3^x)} = e^{x \log 3}$.Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$,wh

Vedclass · Accepted Answer

$\frac{(\log 3)^3}{6}$

Vedclass · Answer

(D) We know that $3^x = e^{\log(3^x)} = e^{x \log 3}$.Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$,where $u = x \log 3$:$3^x = 1 + \frac{x \log 3}{1!} + \frac{(x \log 3)^2}{2!} + \frac{(x \log 3)^3}{3!} + \dots$$3^x = 1 + x \log 3 + \frac{x^2 (\log 3)^2}{2} + \frac{x^3 (\log 3)^3}{6} + \dots$The coefficient of $x^3$ is $\frac{(\log 3)^3}{6}$.

The coefficient of $x^3$ in the expansion of $3^x$ is

Similar Questions

The money invested in a company is compounded continuously. ₹ $400$ invested today becomes ₹ $800$ in $6$ years. What will it become at the end of $33$ years? (Given $\sqrt{2} \approx 1.4142$)

$(1 + 3)\log_e 3 + \frac{1 + 3^2}{2!} (\log_e 3)^2 + \frac{1 + 3^3}{3!} (\log_e 3)^3 + \dots \infty = $

In the expansion of $\frac{a + bx + cx^2}{e^x}$,the coefficient of $x^n$ is

The value of $\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ... + ^n{C_n}}}{{^n{P_n}}}} $ is

The sum of the series $1 + \frac{3}{2!} + \frac{7}{3!} + \frac{15}{4!} + \dots \text{to } \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module