Hello Ally Samaniego,
To work these problems, you will need the following formulas:
(1) $$\sum_{k=1}^n(1)=n$$
(2) $$\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$$
(3) $$\sum_{k=1}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{6}$$
(4) $$\sum_{k=1}^n\left(k^3 \right)=\frac{n^2(n+1)^2}{4}$$
You will also need:
(5) $$\sum_{k=a}^b\left(c\cdot f(k) \right)=c\cdot\sum_{k=a}^b\left(f(k) \right)$$ where $c$ is an arbitrary constant.
(6) $$\sum_{k=a}^b\left(f(k)\pm g(k) \right)=\sum_{k=a}^b\left(f(k) \right)\pm\sum_{k=a}^b\left(g(k) \right)$$
Now, let's look at the problems:
1.) $$S_n=\sum_{i=1}^n(4-9i)$$
Using (6) and (5), we may write:
$$S_n=4\sum_{i=1}^n(1)-9\sum_{i=1}^n(i)$$
Using (1) and (2), we may write:
$$S_n=4n-9\frac{n(n+1)}{2}$$
Combining terms, we obtain:
$$S_n=\frac{8n-9n(n+1)}{2}=\frac{n(8-9(n+1))}{2}=-\frac{n(9n+1)}{2}$$
2.) $$S_n=\sum_{i=1}^n\left((i+1)(i+4) \right)$$
Expanding the summand, we obtain:
$$S_n=\sum_{i=1}^n\left(i^2+5i+4 \right)$$
Using (6) and (5), we may write:
$$S_n=\sum_{i=1}^n\left(i^2 \right)+5\sum_{i=1}^n(i)+4\sum_{i=1}^n(1)$$
Applying (1), (2) and (3), we obtain:
$$S_n=\frac{n(n+1)(2n+1)}{6}+5\frac{n(n+1)}{2}+4n$$
Combining terms, we find:
$$S_n=\frac{n(n+1)(2n+1)+15n(n+1)+24n}{6}= \frac{n\left((n+1)(2n+1)+15(n+1)+24 \right)}{6}= \frac{n\left(2n^2+3n+1+15n+15+24 \right)}{6}= \frac{2n\left(n^2+9n+20 \right)}{6}= \frac{n(n+4)(n+5)}{3}$$
3.) $$S_n=\sum_{i=1}^n\left(i^3-i-2 \right)$$
Using (6) and (5), we may write:
$$S_n=\sum_{i=1}^n\left(i^3 \right)-\sum_{i=1}^n(i)-2\sum_{i=1}^n(1)$$
Using (1), (2) and (4), we have:
$$S_n=\frac{n^2(n+1)^2}{4}-\frac{n(n+1)}{2}-2n$$
Combining terms, we obtain:
$$S_n=\frac{n^2(n+1)^2-2n(n+1)-8n}{4}=\frac{n\left(n(n+1)^2-2(n+1)-8 \right)}{4}=\frac{n\left(n^3+2n^2+n-2n-2-8 \right)}{4}=\frac{n\left(n^3+2n^2-n-10 \right)}{4}$$
4.) $$S_n=\sum_{i=1}^n\left(\frac{3}{n}\left(\frac{i}{n} \right)^2 \right)$$
Using (5), we may write:
$$S_n=\frac{3}{n^3}\sum_{i=1}^n\left(i^2 \right)$$
Using (3), we obtain:
$$S_n=\frac{3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{2n^2}$$
