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

Question

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

anonymous · Answer

$$a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})$$

experimentx · Answer

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

anonymous · Answer

how would I do $$a _{2k-1}+a _{2k}$$

experimentx · Answer

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

experimentx · Answer

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

anonymous · Answer

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