MHB Ally Samaniego's question at Yahoo Answers regarding partial sums

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To find the nth sum of various functions, specific summation formulas are essential. For the first function, the sum of (4-9i) results in S_n = -n(9n+1)/2. The second function, (i+1)(i+4), simplifies to S_n = n(n+4)(n+5)/3 after expansion and applying summation formulas. The third function, (i^3 - i - 2), leads to S_n = n(n^3 + 2n^2 - n - 10)/4. Lastly, the fourth function involving a fraction results in S_n = (n+1)(2n+1)/2n^2. Understanding these steps and formulas is crucial for solving similar problems effectively.
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Here are the questions:

How do I find the nth sum of these functions (calculus I)?


My professor explained it very poorly and I have no idea how to solve these.
If possible, could you walk me through it and give me a good way to think about these to solve them properly? It's greatly appreciated, my exam is next week.

1) Sum of i=1 to n of (4-9i)

2) Sum of i=1 to n of (i+1)(i+4)

3) Sum of i=1 to n of (i^3 - i - 2)

4) Sum of i=1 to n of 3/n(i/n)^2

I don't need all of them done, I just need someone to explain how to do these. I'm so confused. I really appreciate your help!

I have posted a link there to this thread so the OP can see my work.
 
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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:

$$S_n=\frac{3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{2n^2}$$
 
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