1. ## sequence and series

write the formula for the nth term of the sequence
-(1/1), (3/2), -(7/6), (15/24), -(31/120)

2. $(-1)^n \frac{2^n-1}{n!}$

3. $a_1=-\frac{1}{1},...,a_n=(-1)^n\frac{b_n}{c_n}=(-1)^n\frac{b_{n-1} +2^{n-1} }{nc_{n-1}}$

4. Hello, bekah!

Julius is correct!

Write the formula for the $n^{th}$ term of the sequence:
. . $-\frac{1}{1},\;\frac{3}{2},\; -\frac{7}{6},\;\frac{15}{24},\; -\frac{31}{120},\;\hdots$

The signs alternate.
. . This involves: . ${\color{blue}(\text{-}1)^n}$

The numerators are: . $1,\; 3,\; 7,\; 15,\; 31, \hdots$

You might recognize them: one less than a power-of-2.
. . $\begin{array}{ccc}
1 &=& 2^1 - 1 \\ 3 &=& 2^2-1 \\ 7 &=& 2^3 - 1 \\ 15 &=& 2^4-1 \\ 31 &=& 2^5-1 \\ \vdots && \vdots \end{array}$

Hence, the numerator is: . ${\color{blue}2^n-1}$

The denominators are: . $1,\;2,\;6,\;24,\;120,\;\hdots$

You might recognize these as factorials:
. . $\begin{array}{ccc}1 &=&1! \\ 2 &=& 2! \\ 6 &=& 3! \\ 24 &=& 4! \\ 120 &=& ! \\ \vdots & \vdots \end{array}$

Hence, the denominator is: . ${\color{blue}n!}$

Therefore, the $n^{th}$ term is: . $a_n \;=\;(\text{-}1)^n\frac{2^n-1}{n!}$