MHB Find the sum of the first 17 terms of the arithmetic series

[karush]

Find the sum of the first 17 terms of the arithmetic series

$$8+\sqrt{7}, \ 6,\ 4-\sqrt{7}$$
$$u=8+\sqrt{7}$$
$$S_{17} =\frac{u\left(1-\frac{{6}^{17}} {u} \right)}{u}$$

My first shot at this

 

[Greg]

Can you post the question as given?

 

[karush]

Find the sum of the first 17 terms of the arithmetic series

 

[Greg]

Common difference, $d$: $-(2+\sqrt7)$

General term: $a_k=a_1+d(k-1)$

Sum: $\dfrac k2(a_1+a_k)$

 

[MarkFL]

We may write:

$$S_n=a_1+a_2+a_3+\cdots+a_{n-2}+a_{n-1}+a_{n}$$

Since we have an AP, and if we write the common difference as $d$, we may rewrite this as:

$$S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-3)d)+(a_1+(n-2)d)+(a_1+(n-1)d)$$

By the commutative property of addition, we may rewrite this as:

$$S_n=(a_1+(n-1)d)+(a_1+(n-2)d)+(a_1+(n-3)d)+\cdots+(a_1+2d)+(a_1+d)+a_1$$

Adding these two equations, by corresponding terms, we obtain:

$$2S_n=n(2a_1+(n-1)d)$$

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

This is equivalent to the formula given by greg1313.

In the given problem, we know:

$$a_1=8+\sqrt{7},\,d=-(2+\sqrt{7}),\,n=17$$

 

[karush]

$$S_{17}=\frac{17}{2}
\left(2\left(8+\sqrt{7}\right)
+\left(17-1\right)
(-\left(2+\sqrt{7}\right)) \right)$$

Hopefully

 

[MarkFL]

$$S_{17}=\frac{17}{2}
\left(2\left(8+\sqrt{7}\right)
+\left(17-1\right)
(-\left(2+\sqrt{7}\right)) \right)$$

Hopefully

You will want to simplify by distributing/combining like terms, etc. :)

 

[karush]

$S_{17}=\frac{17}{2} \left(2\left(8+\sqrt{7}\right) +\left(17-1\right) (-\left(2+\sqrt{7}\right)) \right)=\sqrt{7}+104$

 

[MarkFL]

I get:

$$S_{17}=-17(8+7\sqrt{7})$$