Question

I am trying to find the nth of a sequence : a. 1, 3, 6, 10, 15 ..i need to know how to do this

Answer 1

what value of n are we trying to find?
in other words, which term are you trying to find?

Answer 2

looks like you are adding one more each time. or you can figure out an equation

Answer 3

1+2=3
3+3=6
6+4=10
10+5=15
etc so next one would be
15+6=21
21+7=28
etc

Answer 4

This is the question..What is the nth term in each of the numerical patterns.and those are the numbers

Answer 5

I don't need to know what comes next, but a formula

Answer 6

\[a_{n}=a_{n-1}+n\]

Answer 7

johnnyrocket has it but there is a closed form as well
i will see if i can find it

Answer 8

If we are looking for a general formula for the sequence it would be
\[a _{n}=a _{n-1}+n\] for all n>1
where\[n _{1}=1\]

Answer 9

i have three differnet sequences so i need to learn how to plug in the numbers or me uwith a fomrula

Answer 10

i think we want the closed form of this thing.

Answer 11

i am getting there i think. give me a few

Answer 12

((n+1)n)/2

Answer 13

(n^2+n)/2

Answer 14

i believe you but why the heck to i keep getting
\[\frac{n^2-n}{2}\] which i know is wrong!

Answer 15

maybe bad algebra? or bad method.

Answer 16

once you have it it is clear. for some reason i kept getting minus sign
can you show me how to get it right?

Answer 17

How do you get a negative

Answer 18

oh now i see it. i was summing
\[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] when i should have been summing
\[\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}\]

Answer 19

then
\[n^2-\frac{n(n-1)}{2}\]will work. anyway that is what i did. was there a snappier way?

Answer 20

my way was probably the donkey way. i am sure there is a snap way to do this, i forget how

Answer 21

believe it or not i just saw this a couple years ago and now totally forget. solve these and recursions like
\[a_n=a_{n-1}+2a_{n-2}\] etc

Answer 22

i would love to see what you did to get this