I remember doing a question like this before but I forgot what to do
Follow Math Help Forum on Facebook and Google+
$\displaystyle{\sum_{k=0}^{1008}}~(2k+1) = $ $2\displaystyle{\sum_{k=0}^{1008}}~k+\displaystyle {\sum_{k=0}^{1008}}~1 = $ $2 \dfrac{(1008)(1009)}{2} + 1009 = 1009(1008+1) = 1009^2 = 1018081$
$a_1 + (a_1+1) + (a_1+2) + \, ... \, + (a_1+2017) = 27243$ $S_{2018} = 27243 = \dfrac{2018}{2}[a_1 + (a_1+2017)] \implies a_1 = -995$ sum of odd terms ... $-995 + (-993) + \, ... \, + (-995+2016) = \dfrac{1009}{2}(-995 + 1021) = 13117$
these answers are conflicting
Originally Posted by Ilikebugs these answers are conflicting The method and answer in post #3 are the correct ones.