<s>[INST] Compute $\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$ Compute $$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$$
$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$

$k - 2 = t , k = t + 2$

$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^{t+2}}{3^{2t+1}} - \binom {n}{n} \frac{2^n}{3^{2n}}$$ 
Let's simplify this:  
$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^{t+2}}{3^{2t+1}} =  \frac{4}{3}\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}}$$
Now let's calculate this: $\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}}$

$a = 1 , b = 2/9$

$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}} = (1+\frac{2}{9})^n$$
My final answer: $\frac{4}{3}(\frac{11}{9})^n - \frac{2^n}{3^{2n
}}$
Opinions?
 [/INST] Let $k-2=u$
$$=3\sum_{u=-1}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^{u+2}=3\left(\dfrac2{3^2}\right)^2\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u$$
Now,
$$\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u=\left(1+\dfrac2{3^2}\right)^n-\binom nn\left(\dfrac2{3^2}\right)^n=?$$
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