Question

find the sum of the first 50 terms of the sequence.. a(n)=3n+2
do I just plug in 50? so the answer is 152?

Answer 1

you need the sum

3(1) + 2 + 3(2) + 2 + 3(3) + 2

Answer 2

no, that would be the 50th term, not the sum of all 50 terms

Answer 3

You have to find the sum of all 50 elements
{5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, \
56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, \
107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, \
149, 152}

Answer 4

Of course, you can add them term by term or you can think about a shorter way to do it

Answer 5

i like your way better :)

Answer 6

Think about
\[
1+2 +3 \cdots + (n-1)+n = \frac {n(n+1)}2
\]

Answer 7

\[
\sum_{n=1}^{50} 3 n +2 = 2(50 )+ 3 \frac {50* 51}2=3925

\]

Answer 8

@hi5tigtg58 , but how do you do this sum for the first 10,000 elements. You are not going to add them up term by term.

Answer 9

Yes, and also about \[\sum_{1}^{n}1=n\]

If
\[a ^{n}=3n+2\]
then the sum of the first n terms of this series could be expressed as \[\sum_{1}^{n}(3n+2)=3\sum_{1}^{n}n+2\sum_{1}^{n}1\]

Answer 10

So, Hi-5, what would the value of n be in this case?
How would you evaluate \[\sum_{1}^{n}n? \sum_{1}^{n}1?\]for that n?

Answer 11

(That's two separate sums.)