Compute $\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$

Question

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?

Answer 1

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, $$\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u=\left(1+\dfrac2{3^2}\right)^n-\binom nn\left(\dfrac2{3^2}\right)^n=?$$