Question

Find the first, fourth, and eighth terms of the sequence.
A(n)=-2*2^n-1

Answer 1

hmmm \(\large \bf a_{\color{brown}{ n}}=2\cdot 2^{{\color{brown}{ n}}-1}\)

Answer 2

\(\large \bf a_{\color{brown}{ n}}=2\cdot 2^{{\color{brown}{ n}}-1}?\) rather...is that... the formula

Answer 3

\(\large {
a_{\color{brown}{ n}}=2\cdot 2^{{\color{brown}{ n}}-1}
\\ \quad \\

\begin{array}{llll}
term&value
\\\hline\\
a_{\color{brown}{ 1}}&a_{\color{brown}{ 1}}=2\cdot 2^{{\color{brown}{ 1}}-1}\to 2\cdot 2^0\to 2\cdot 1\to 2\\

a_{\color{brown}{ 4}}&a_{\color{brown}{ 4}}=2\cdot 2^{{\color{brown}{ 4}}-1}\to 2\cdot 2^3\\

a_{\color{brown}{ 8}}&a_{\color{brown}{ 8}}=2\cdot 2^{{\color{brown}{ 8}}-1}\to 2\cdot 2^7\\
\end{array}
}\)

Answer 4

You can do this to make this problem easier: \(\Large 2\cdot2^{n-1}\rightarrow 2^1\cdot2^{n-1}\rightarrow 2^{n-1}+1=\boxed{2^n}\)

Answer 5

\(\Large 2^{n-1+1}\), not \(\Large 2^{n-1}+1\)

Answer 6

Thanks.