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[SOLVED] sequences

dwsmith

Well-known member
Feb 1, 2012
1,673
Use a geometric or algebraic argument to find a formula for the partial sums $A_n$ of an arithmetic sequence.

I know that the partial sum is $S_n = n/2(2a_1+(n-1)d)$ where d is the difference.

$A_n = \sum\limits_{k = 1}^n a_k$

I can come up with $n/2(a_1+a_n)$ but how do I get the difference?
 

soroban

Well-known member
Feb 2, 2012
409
Hello, dwsmith!

Use a geometric or algebraic argument to find a formula for the partial sum $S_n$ of an arithmetic sequence.

I know that the partial sum is: $S_n = \frac{n}{2}(2a_1+(n-1)d)$ where d is the difference.

$A_n = \sum\limits_{k = 1}^n a_k$

I can come up with $\frac{n}{2}(a_1+a_n)$ but how do I get the difference?

Why do you want $d$ ? .You don't know $a_1$ either.

Did you read the question?
They ask us to derive the partial sum formula.


We have: .[tex]\begin{Bmatrix}a_1 &=& a_1 \\ a_2 &=& a_1 + d \\ a_3 &=& a + 2d \\ a_4 &=& a+3d \\ \vdots && \vdots \\ a_n &=& a + (n\!-\!1)d \end{Bmatrix}[/tex]

Add: .[tex]\underbrace{a_1 + a_2 + a_3 + \cdots + a_n}_{S_n} \;=\;\underbrace{a_1 + a_1 + \cdots + a_1}_{n\text{ terms}} + \big[1 + 2+ 3+\cdots + (n\!-\!1)\big]d [/tex]

. . . . . . . [tex]S_n \;=\;n\!\cdot\!a_1 + \frac{n(n-1)}{2}d \;=\;\frac{2na_1 + n(n\!-\!1)d}{2}[/tex]

. . . . . . . [[tex]S_n \;=\;\frac{n}{2}\big[2a_1 + (n\!-\!1)d\big][/tex]
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The nth term of an arithmetic sequence is:

$\displaystyle a_n=a_1+(n-1)d$

We could derive the partial sum by expressing the sum recursively:

$\displaystyle S_{n}=S_{n-1}+(n-1)d$

$\displaystyle S_{n+1}=S_{n}+nd$

Subtracting the former from the latter, we find:

$\displaystyle S_{n+1}=2S_{n}-S_{n-1}+d$

$\displaystyle S_{n+2}=2S_{n+1}-S_{n}+d$

Subtracting again, we find the homogeneous recursion:

$\displaystyle S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}$

The associated characteristic equation is:

$\displaystyle r^2-3r^2+3r-1=0$

$\displaystyle (r-1)^3=0$

Hence:

$\displaystyle S_n=k_1+k_2n+k_3n^2$

Using:

$\displaystyle S_1=k_1+k_2+k_3=a_1$

$\displaystyle S_2=k_1+2k_2+4k_3=2a_1+d$

$\displaystyle S_3=k_1+3k_2+9k_3=3a_1+3d$

we find:

$\displaystyle k_1=0,k_2=a_1-\frac{1}{2}d,k_3=\frac{1}{2}d$

And so:

$\displaystyle S_n=\left(a_1-\frac{1}{2}d \right)n+\left(\frac{1}{2}d \right)n^2$

$\displaystyle S_n=\frac{n}{2}\left(2a_1-d+nd\right)$

$\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$