Compute $\sum_{k=4}^{n=30} \frac{C(25,k-3) * 4^{2k-7}}{k-2}$

Question

Compute $\sum_{k=4}^{n=30} \frac{C(25,k-3) * 4^{2k-7}}{k-2}$

I started by saying $t = k-3$, so $k=t+3$ then I got 4$\sum_{k=1}^{n=27} C(25,t) * \frac{4^t}{t+1}$ and then

$\sum_{k=4}^{n=27} C(27,t) * \frac{4^t}{(t+1)26*27}$

and now I'm stuck getting rid of $1/(t+1)$, if it was $(t+1)$ I could differntiate and solve it, I believe. but if i learned how to integrate than I believe I would know how to get rid of $1/(t+1)$ , and we just started studing integrals in calculus so I'm pretty weak in that part.

What do you guys think?

Answer 1

I get to: $(0.25\sum_{t=0}^{n=25} \frac{C(25,t) * 16^{t}}{1+t}) -0.25$

and from here I use binom's formula and write $(1+x)^n$, integrate it and the equivalent binom's formula and get to the final answer: $(0.25 * 17^{26} / 26) -0.25$