Question

Find the sum of the first 30 terms of the sequence below.
an=4n+1

Answer 1

an=4(1)+1=5
then you do an=4(2)+1=9
then so on and so forth until it is an=4(30)+1=121
good luck

Answer 2

you understand?

Answer 3

there is a better way

Answer 4

that will take forever

Answer 5

First term is: 4(1) + 1
Second term is: 4(2) + 1
Third term is: 4(3) + 1
...
...
30th term is: 4(30) + 1

Add them all up: 4(1 + 2 + 3 + ..... + 30) + (1 + 1 + 1 ... 30 times)
The first parenthesis is the sum of the first 30 natural numbers. There is a standard formula for that: n(n+1) / 2 where n = 30
The second parenthesis has the same number 1 repeating 30 times.
Now you can addd them up.

Answer 6

the 1 is repeated 30 time so the sum of number 1 is 30

and we know that the \[\sum_{1}^{ n} n = [n*(n+1)]/2\]
so we have
4*[n*(n+1)/2]+30
and the n is 30

Answer 7

understand??

Answer 8

the really easy method is

\[S_{30} = \frac{30}{2}\times[5 + 121]\]

Answer 9

the sequence is arithmetic, the 1st term is 5 and the last term is 121.

so add the 1st and last terms then multiply by 15, half the number of terms

Answer 10

1890 is the answer?

Answer 11

Yes, 1890.

Answer 12

the formula for 1st and last terms is

\[S_{n} = \frac{n}{2} \times (a + l)\]

a = 1st term, l = last term and n = number of terms

Answer 13

yes, thats the answer, well done

Answer 14

thank you