We have
[tex]2+\dfrac54+\dfrac{11}{16}+\dfrac{23}{64}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{3\cdot2^n-1}{4^n}[/tex]
(notice that each denominator is a power of 4, and each numerator is one less than some multiple of 3, in particular 3 times some power of 2)
Recall for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
So we have
[tex]\displaystyle\sum_{n=0}^\infty\frac{3\cdot2^n-1}4=3\sum_{n=0}^\infty\left(\frac12\right)^n-\sum_{n=0}^\infty\left(\frac14\right)^n=\frac3{1-\frac12}-\frac1{1-\frac14}=\boxed{\frac{14}3}[/tex]
