Adding Every Other Odd

[AlejandroVerd]

So, I'm having problems figuring this out, and it's driving me crazy because I know I'm most likely just over-thinking the problem.

So, I'm just trying to figure out a formula I can use to add every other odd number (S1=1, S2=1+5, S3=1+5+9)

Thanks for the help.

 

[tkhunny]

That sounds like a delightful exploration.

Have you consdiered adding up ALL the odd numbers and subtracting half of them?
Have you considered that each of the ones you want to sum are exactly two less that each of it's nearest greater neighbor in the odds?

 

[soroban]

Hello, AlejandroVerd!

Here's one way to solve it . . .

Find a formula to add every other odd number.
. . \(\displaystyle S_1 = 1,\;\; S_2 = 1+5,\;\; S_3 = 1+5+9,\;\hdots\)

Consider the sequene formed by the sums.
Take the differences of consecutive terms, then take differences of the differences, and so on.

. . \(\displaystyle \begin{array}{ccccccccccccccc}S_n & 1 && 6 && 15 && 28 && 45 && 66 && \hdots \\ \text{1st diff} && 5 && 9 && 13 && 17 && 21 && \hdots \\ \text{2nd diff} &&& 4 && 4 && 4 && 4 && \hdots\end{array}\)


Note that the second differences are constant.
This tells us that the generating function of the second degree . . . a quadratic.

The general quadratic form is: .\(\displaystyle S(n) \:=\:an^2 + bn + C\)


Use the first three terms of the sequence to determine \(\displaystyle a,b,c.\)

. . \(\displaystyle \begin{array}{ccccc}S(1) = 1\!: & a + b + c &=& 1 & [1] \\ S(2) = 6\!: & 4a + 2b + c &=& 6 & [2] \\ S(3) = 15\!: & 9a + 3b + c &=& 15 & [3] \end{array}\)

\(\displaystyle \begin{array}{ccccc}\text{Subtract [2]-[1]:} & 3a + b &=& 5 & [4] \\ \text{Subtract [3]-[2]:} & 5a + b &=& 9 & [5] \end{array}\)

\(\displaystyle \begin{array}{ccccccc}\text{Subtract [5]-[4]:} & 2a \:=\:4 & \Rightarrow & a \,=\,2 \end{array}\)

\(\displaystyle \begin{array}{ccccccc}\text{Substitute into [4]:} & 3(2) + b \:=\:5 & \Rightarrow & b \:=\:-1 \end{array}\)

\(\displaystyle \begin{array}{ccccccc}\text{Substitute into [1]:} & 2 - 1 + c \:=\:1 & \Rightarrow & c \:=\:0 \end{array}\)


\(\displaystyle \text{Therefore, the generating function is: }\;S(n) \;=\;2n^2 - n \;=\;n(2n-1)\)

 

[Denis]

Hmmm....seems to me you simply want 1 + 5 + 9 + 13 + 17 + 21 + 23 ...
f = first term
d = difference
n = number of terms

formula (which you should know!): sum = n[2f + d(n-1)] / 2

If n = 5:
sum = 5[(2(1) + 4(5-1)] / 2 = 45

 

[lookagain]

... a formula I can use to add every other odd number (S1=1, S2=1+5, S3=1+5+9)

Partially alternate way:

Let n = the number of terms.

The differences are 4, and based on the first term, the general term is (4n - 3).

Write the descending sum under the ascending sum and add the two equations
together to get twice the sum on one of the sides. Then divide each side of
the resulting equation by 2:


\(\displaystyle S \ = \ 1 \ \ \ \ \ \ \ \ \ + \ 5 \ \ \ \ \ \ \ \ \ + \ 9 \ \ \ \ \ \ \ \ \ + ... + (4n - 7) + (4n - 3)\)
\(\displaystyle S \ = \ (4n - 3) + (4n - 7) + (4n - 11) + ... + 5 \ \ \ \ \ \ \ \ \ + 1\)
----------------------------------------------------------------------------------------------
\(\displaystyle 2S = \ (4n - 2) + (4n - 2) + (4n - 2) \ + ... + (4n - 2) + (4n - 2)\)


\(\displaystyle S \ = \ (2n - 1) + (2n - 1) + (2n - 1) \ + ... + (2n - 1) + (2n - 1)\)


There are n terms, which makes n columns, so the sum is:

\(\displaystyle S \ = \ n(2n - 1)\)

 

[Denis]

\(\displaystyle S \ = \ n(2n - 1)\)

To students who may look at this:
Above formula to be used ONLY if first term is 1.
Will not work if, as example, sequence is 3,7,11,15...

 

[lookagain]

So, I'm just trying to figure out a formula I can use to
\(\displaystyle > \ >\\)add every other odd number (S1=1, S2=1+5, S3=1+5+9)\(\displaystyle < \ <\)

\(\displaystyle S \ = \ n(2n - 1)\)

To students who may look at this:
Above formula to be used ONLY if first term is 1.
Will not work if, as example, sequence is 3,7,11,15...

That is true, but that's not an issue here because the problem was specifically to add every other odd number,
starting with the odd number 1, as the problem did not state to come up with a formula for just any set
composed of every other odd numbers.

Note: The formula for adding every other odd number (S1 = 3, S2 = 3 + 7, S3 = 3 + 7 + 11, ...)
turns out to be n(2n + 1).

 

[Denis]

No need to explain.
I simply wanted to help the "student".
Thank you.