<s>[INST] ${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100 \cdot6\over 4}$ Find the value of
$${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100\cdot6\over 4}$$
My approach:
$${2\over 4^{100}}+{2\over 4^{99}}+\cdots+{2\over 4^{1}}$$
forms a G.P.  But how to separate $\displaystyle{6\over 4^{100}}+{2\cdot6\over 4^{99}}+\cdots$?
 [/INST] The second piece is simply
$$6 \sum_{k=1}^{100} \frac{k}{4^{101-k}}= \frac{6}{4^{101}} \sum_{k=1}^{100} k \, 4^k$$
To evaluate, let $S = 4 + 2 \cdot 4^2 +3 \cdot 4^3 + 4 \cdot 4^4+\cdots+100\cdot 4^{100}$  Then
$$4 S = 4^2+ 2 \cdot 4^3 +3 \cdot 4^4 + 4 \cdot 4^5+\cdots+100 \cdot 4^{101}$$
$$\implies S-4S=-3 S = 4+4^2+4^3+4^4+\cdots+4^{100}-100\cdot 4^{101}$$
Now you have another geometric series, and use that fact to solve for $S$.
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