Sequences and Series

If the first term of an arithmetic sequence is 3 and the common difference is 2, find the sum of the first 10 terms.

Find the sum of the geometric sequence \(3, 6, 12, 24, \ldots\) up to the nth term.

Consider the sequence \(a_n\) where \(a_1 = 1\), \(a_2 = 2\), and for \(n \geq 3\), \(a_n = 2a_{n-1} - a_{n-2}\). Find the next term \(a_6\) of the sequence?

Given a sequence of numbers: 2, 5, 10, 17, find the formula for the nth term and calculate the 8th term of the sequence.

Find the 10th term of the sequence defined by \( a_n = 2n^2 + 3n + 1 \)

Find the sum of the first 100 terms of the sequence defined by \(a_n = n^2\).

Question 10 ( 4 points) Let \( g(x)=x \cos (3 x) \). Find the Maclaurin series of its derivative, \( g^{\prime}(x) \) None of the other answers are correct. \( 3 x-\frac{9}{2} x^{3}+\frac{81}{40} x^{5}+\cdots \) \( 3-\frac{3}{2} x^{2}+\frac{5}{8} x^{4}+\cdots \) \( 1-\frac{27}{2} x^{2}+\frac{135}{8} x^{4}+\cdots \) \( 3-\frac{27}{2} x^{2}+\frac{81}{8} x^{4}+\cdots \)

Find the indicated sum. \[ \sum_{S_{7}=1}^{7} 5^{k} \]

Find the sum of $\sum_{n=0}^{\infty} \frac{n(n+1)}{3^{n-1}}$ by identifying it as the value of the derivative of the series \[ g(x)=\sum_{n=0}^{\infty}(n+1) x^{n}=\frac{1}{(1-x)^{2}} \text { for }|x|<1 . \] a. Take the derivative of the series given by $g(x)$. Write the derivative of the series in the first box and the derivative of the rational expression in the second box. \[ g^{\prime}(x)=\sum_{n=1}^{\infty} \square= \] b. Find the value of $g^{\prime}\left(\frac{1}{3}\right)=\sum_{n=0}^{\infty} \frac{n(n+1)}{3^{n-1}}$. \[ g^{\prime}\left(\frac{1}{3}\right)= \]

For the sequence, determine if the divergence test applies and either state that $\lim _{n \rightarrow \infty} a_{n}$ does not exist or find $\lim _{n \rightarrow \infty} a_{n}$. (If an answer does not exist, enter DNE.) \[ \begin{array}{r} a_{n}=\frac{2^{n}+3^{n}}{10^{n / 2}} \\ \lim _{n \rightarrow \infty} a_{n}= \end{array} \]

Test the series below for convergence using the Ratio Test. \[ \sum_{n=1}^{\infty} \frac{n^{4}}{1.5^{n}} \] The limit of the ratio test simplifies to $\lim _{n \rightarrow \infty}|f(n)|$ where \[ f(n)= \] The limit is: (enter oo for infinity if needed) Based on this, the series Select an answer $\checkmark$

Find the series' radius of convergence. \[ \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{\ln (n+1)} \] A) 1 B) $\infty$, formall $x$ C) 0 D) 2