Question

Help !
I am trying to solve this series with out using a calculator. Its been a while since Ive done these.
my approach was
a_nth term = first term + difference from first term to second *(nth term - 1) ....

Answer 1

can some one help me interpret the solution ?

Answer 2

I got for the left term 11/2 , easy . but the right term is difficult

Answer 3

as far as the solution goes , I understand they use eulers form to express cosine

Answer 4

Hint : \[\cos x = \mathfrak{ R} ~e^{ix}\]

Answer 5

expand the given sum
you will see why it is geometric

Answer 6

\[\sum_{-10}^{0} 1/2 e^ (j3\pi/8)\]

Answer 7

\[\begin{align}\sum\limits_{k=-10}^0 e^{j3\pi k/8} &= e^{-j10*3\pi/8} +e^{-j9*3\pi/8} +e^{-j8*3\pi/8} +\cdots + e^{-i0*3\pi/8} \end{align}\]

Answer 8

\[\begin{align}\sum\limits_{k=-10}^0 e^{j3\pi k/8} &= e^{-j10*3\pi/8} +e^{-j9*3\pi/8} +e^{-j8*3\pi/8} +\cdots + e^{-i0*3\pi/8}\\~\\
&= (e^{-j*3\pi/8})^{10} +(e^{-j*3\pi/8})^9 +(e^{-j*3\pi/8})^8 +\cdots + (e^{-i*3\pi/8})^0\\~\\
\end{align}\]

Answer 9

see the geometric series yet ?

Answer 10

of course.... I am refering to my calc book to refer to geo series.. one sec

Answer 11

\[\sum_{}^{} a*r ^{n-1}\]