Sigma notation help: 2 + 5 + 8 + 11 + ... + 29

[yummymummy1713]

I need to write this in sigma notation. I know that it is going up by three, but I can't think of anything. Do you have any ideas.

2 + 5 + 8 + 11 + ... + 29

 

[stapel]

You have found the difference, 3. That's good.

What is the first term? Can you relate this to "n", the sigma-notation counter ("index")?

. . . . .n = 1: a_n = 2
. . . . .n = 2: a_n = 5
. . . . .n = 3: a_n = 8
. . . . .n = 4: a_n = 11

Can we put that "add another 3" thing into account?

. . . . .n = 1: a_n = 2
. . . . .n = 2: a_n = 2 + 3
. . . . .n = 3: a_n = 2 + 3 + 3
. . . . .n = 4: a_n = 2 + 3 + 3 + 3

In other words:

. . . . .n = 1: a_n = 2 + 0(3)
. . . . .n = 2: a_n = 2 + 1(3)
. . . . .n = 3: a_n = 2 + 2(3)
. . . . .n = 4: a_n = 2 + 3(3)

Note that each "(this) times 3" thing is one less than the index value for that step. Figure out a way to relate this to n, and you'll have your formula for the n-th term, a_n.

Once you have this formula, plug "29" in for "a_n", and solve for n. This will tell you how many terms you have in your summation, starting with n = 1 and going up to whatever value you just solved for.

Eliz.

 

[soroban]

Hello, Yummy!

Write in sigma notation: $\displaystyle \,2 \,+\,5\,+\,8\,+\,11\,+\,\cdots\,+\,29$

I know that it is going up by three. . . Good start!

A function that "goes up by three" would be: $\displaystyle \,f(n)\:=\:3n$

But this gives us: $\displaystyle \:3,\,6,\,9,\,12,\,\cdots\,30$
. . . And we want: $\displaystyle \,2,\,5,\,8,\,11,\,\cdots\,29$

Got any ideas?