1. sum of integers

Find the sum of the first 100 positive even integers. Explain how this sum can be found quickly.
Sn = (n/2)(a1 + an)
is that the correct formula?

2. Originally Posted by aikenfan
Find the sum of the first 100 positive even integers. Explain how this sum can be found quickly.
Sn = (n/2)(a1 + an)
is that the correct formula?
that is the correct formula. note that we can consider an ordered list of even integers as an arithmetic sequence. we use that formula to find the sum of the terms of an arithmetic sequence. you know that $a_1 = 2$, and we want $n = 100$, so now the only thing left for you to find is $a_100$

3. Hello, aikenfan!

Find the sum of the first 100 positive even integers.
Explain how this sum can be found quickly.

Sn = (n/2)(a1 + an)
is that the correct formula?
That formula will work, but I prefer the general Sum Formula:

. . $S_n \;=\;\frac{n}{2}\left[2a_1 + (n-1)d\right]$

It is more complicated, but we don't need to calculate the 100th term.

We have: . $a_1 = 2,\;d = 2,\;n = 100$

Therefore: . $S_{100} \;=\;\frac{100}{2}\left[2\!\cdot\!2 + (100-1)2\right] \;=\; 10,100$

4. I am not sure how to calculate the 100th term?
an = dn + c?

5. Originally Posted by aikenfan
I am not sure how to calculate the 100th term?
an = dn + c?
you would use the formula $a_n = a_1 + (n - 1)d$ (which is the general formula for the nth term of an arithmetic sequence) where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the current number of the term, and $d$ is the common difference. you would use the values Soroban gave

thus to find $a_{100}$ you would compute $a_{100} = 2 + (100 - 1)(2)$

6. an = dn + c

Isn't this the same thing? I get 200 either way...

7. Originally Posted by aikenfan
an = dn + c

Isn't this the same thing? I get 200 either way...
well, if you define c as a_1 - d, then yes. but i've never seen anyone use "c" in these formulas

8. Sn = (100/2)(2+200) = 10100