Two Plus Six Plus Another Then Plus Another

Here we observe that the general term is $\quad\displaystyle\sum_{n=1}^{100}\frac{2+6n}{4^{101-n}}$

Hence it can be rewritten as $:\quad\frac{1}{4^{101}} \displaystyle(2\sum_{n=1}^{100}4^n+6\sum_{n=1}^{100}n4^n)$

$1^{st}$ term is a simple GP and $2^{nd}$ term is Arithmetico-Geometric progression Hence applying respective formulas of GP and AGP we get

$\displaystyle:\quad\frac{1}{4^{101}}(\frac{8(4^{100}-1)}{3}+\frac{8(1+299\times4^{100})}{3})$ On simplifying answer is $200$

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Sum of AGP of the form $\displaystyle:\sum_{k=1}^{n}[a+(n-1)d]r^{k-1}=a+[a+d]r+[a+2d]r^2.....+[a+(n-1)d]r^{n-1}$
is $\displaystyle S_n=\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r)^2}$

On substituting given values we can compute the above sum.

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Here we observe that the general term is $\quad\displaystyle\sum_{n=1}^{100}\frac{2+6n}{4^{101-n}}$ Hence it can be rewritten as $:\quad\frac{1}{4^{101}} \displaystyle(2\sum_{n=1}^{100}4^n+6\sum_{n=1}^{100}n4^n)$ we know that $\sum_{n=1}^{100}x^n= \dfrac{x^{101}-x}{x-1}$ deriving both sides, and multiplying by x, $\sum_{n=1}^{100}nx^n=\dfrac{100x^{102}-101x^{101}+x}{(x-1)^2}$ entering $x=4$ , $\dfrac{1}{4^{101}}(2*\dfrac{4^{101}-1}{3}+6*\dfrac{100*4^{102}-101*4^{101}+4}{9}$ upon some simplificaton, the answer is $\boxed{200}$

pattern tn = (2+i*6)/(4^(101-i)) i= 1,2,3....,100