设$S=sumlimits_{k=1}^{+infty}[dfrac{116+3^{k-1}}{3^k}]\
T=sumlimits_{k=1}^{+infty}[dfrac{116+2*3^{k-1}}{3^k}]\$
则S+T=_____
提示:由埃尔米特恒等式:$[x]+[x+dfrac{1}{n}]+cdots+[x+dfrac{n-1}{n}]=[nx]$
故$sumlimits_{k=1}^{+infty}left([dfrac{116}{3^k}+dfrac{1}{3}]+[dfrac{116}{3^k}+dfrac{2}{3}]
ight)
=sumlimits_{k=1}^{+infty}left([dfrac{116}{3^{k-1}}]-[dfrac{116}{3^k}]
ight)=[dfrac{116}{3^0}]=116$