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?
could someone please help me?
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 $\displaystyle a_1 = 2$, and we want $\displaystyle n = 100$, so now the only thing left for you to find is $\displaystyle a_100$
Hello, aikenfan!
That formula will work, but I prefer the general Sum Formula: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?
. . $\displaystyle S_n \;=\;\frac{n}{2}\left[2a_1 + (n-1)d\right]$
It is more complicated, but we don't need to calculate the 100th term.
We have: .$\displaystyle a_1 = 2,\;d = 2,\;n = 100$
Therefore: .$\displaystyle S_{100} \;=\;\frac{100}{2}\left[2\!\cdot\!2 + (100-1)2\right] \;=\; 10,100$
you would use the formula $\displaystyle a_n = a_1 + (n - 1)d$ (which is the general formula for the nth term of an arithmetic sequence) where $\displaystyle a_n$ is the nth term, $\displaystyle a_1$ is the first term, $\displaystyle n$ is the current number of the term, and $\displaystyle d$ is the common difference. you would use the values Soroban gave
thus to find $\displaystyle a_{100}$ you would compute $\displaystyle a_{100} = 2 + (100 - 1)(2)$