Question

very desperate!

find the sum of the first 50 terms of the sequence an=3n+2

Answer 1

\[1+2+\cdots+k=\frac{k(k+1)}{2} \\ \text{ multiplying both sides by a constant } c \text{ gives } \\ c(1+2+\cdots+ k)=c \cdot \frac{k(k+1)}{2} \\ \\ \text{ also if you are adding an } k \text{ amount of two's together } \\ \text{ like } 2+2+2+ \cdots +2 =2k\]

Answer 2

so we can look at this seperate
you have
\[a_n=b_n+c_n \\ \sum a_n=\sum(b_n+c_n)=\sum b_n +\sum c_n\]
that big looking weird symbol just means the sum of the terms
\[\text{ anyways } 1+2+3 +\cdots +50=?\]

Answer 3

how do i plug the sequence into the sigma?

Answer 4

or the equation?

Answer 5

\[\sum_{n=1}^{50}(3n+2)=\sum_{n=1}^{50} (3n)+\sum_{n=1}^{50}2 =3 \sum_{n=1}^{50} n +\sum_{n=1}^{50} 2\]
so this is why I'm asking you to first calculate 1+2+3+...+50
I have given you a formula above to this

Answer 6

this notation I wrote
\[\sum_{n=1}^{50}(3n+2)=\sum_{n=1}^{50} (3n)+\sum_{n=1}^{50}2 =3 \sum_{n=1}^{50} n +\sum_{n=1}^{50} 2 \\ \text{ can be written as } \\ 3(1+2+3 + \cdots +50)+(2 +2 + \cdots +2)\]
that is suppose to be fifty amount of twos

Answer 7

is it 3925?

Answer 8

that is right
\[3 \cdot \frac{50(51)}{2}+2(50) \text{ using both of the formulas I mentioned \in the first post }\]

Answer 9

oh okay! so that's the answer t the question?

Answer 10

what I mean by " that is right" is "that is correct"

Answer 11

thank you!