Compute sum to n

Question

I'm confused as to how to finish answering this question.

Compute $$\sum_{i=28}^n (3i^2 - 4i + \frac{5}{7^i}) $$

I end up with $$ 3[\frac{n(n+1)(2n+1)}{6}] - 4[\frac{n(n+1)}{2}] + [\frac{k^n(5(-log(7))^n}{n!}]$$

I'm not entirely sure if the last part of my answer is right but my main question is what do I do about the $ i=28$? Any help would be awesome. Thanks

Answer 1

$$\sum_{i=28}^n \left(3i^2 - 4i + \frac{5}{7^i}\right) =\sum_{i=1}^n \left(3i^2 - 4i + \frac{5}{7^i}\right) - \sum_{i=1}^{27} \left(3i^2 - 4i + \frac{5}{7^i}\right) = 3\left(\frac{n(n+1)(2n+1)}{6}\right) - 4\left(\frac{n(n+1)}{2}\right) + 5\frac{\left((1-\left(\frac{1}{7}\right)^{n+1}\right)}{\frac{6}{7}}-\left(3\left(\frac{27(27+1)(2\cdot 27+1)}{6}\right) - 4\left(\frac{27(27+1)}{2}\right) + 5\frac{\left((1-\left(\frac{1}{7}\right)^{27+1}\right)}{\frac{6}{7}}\right) = \\ n^3-\frac{n^2}{2}-\frac{3 n}{2}-\frac{(5\cdot 7^{-n})}{6}+\frac{35}{6} - \frac{1267186243758001135830923954}{65712362363534280139543}= \\ n^3-\frac{n^2}{2}-\frac{3 n}{2}-\frac{(5\cdot 7^{-n})}{6}-\frac{7600817529865283115180659719}{394274174181205680837258}$$

Answer 2

For $n\ge28,$

$$\sum_{i=28}^n \left(3i^2 - 4i + \frac{5}{7^i}\right) $$

$$=\sum_{i=1}^n \left(3i^2 - 4i + \frac{5}{7^i}\right) -\sum_{i=1}^{27} \left(3i^2 - 4i + \frac{5}{7^i}\right) $$

Now we can safely apply the formula you have specified