Question

a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})
for a positive interger k, find the simplified expression for \[a _{2k-1}\]

Answer 1

\[a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})\]

Answer 2

\[ a _{2k - 1}=((-1)^{2k-2}(6)+2)/(4k-1+(-1)^{2k-2}) \]

Answer 3

how would I do \[a _{2k-1}+a _{2k}\]

Answer 4

\[ a _{2k-1}=((-1)^{2k-2}(6)+2)/(4k-1+(-1)^{2k-2}) \]
\[ a _{2k}= ((-1)^{2k-1}(6)+2)/(4k+1+(-1)^{2k-1}) \]
\[ \frac{((-1)^{2k-2}(6)+2)}{(4k-1+(-1)^{2k-2})} + \frac{((-1)^{2k-1}(6)+2)}{(4k+1+(-1)^{2k-1})}\]

\[ \frac{6+2}{4k-1+1} + \frac{-(6)+2}{(4k+1-1)}\]

Answer 5

(-1)^2k-2 is +1
(-1)^2k-1 is -1

Answer 6

are they not both over 4k and can be added easily?