Question 5edfb

Dec 7, 2017

$10923$
See explanation.

Explanation:

So you have a sequence that goes:
3, 8, 13, 18,...

One possibility that I see right away is that $8 - 3 = 5$
then $13 - 8 = 5$, then $18 - 13 = 5$, so perhaps the sequence is just adding 5 to the previous number, starting from 3.

If we call the first entry, $3 = {k}_{1}$,
and the second entry of sequence, $8 = {k}_{2}$,
we can rewrite the first and second entries as follows:
${k}_{1} = 3$
${k}_{2} = {k}_{1} + 5$
the third entry would be written:
${k}_{3} = {k}_{2} + 5$
fourth entry:
${k}_{4} = {k}_{3} + 5$
...
sixty-sixth entry:
${k}_{66} = {k}_{65} + 5$

We want the sum of ${k}_{1}$ to ${k}_{66}$, that is, the sum of all the equations shown above.
We now notice that we can rewrite these equations as such:
${k}_{1} = 3$
${k}_{2} = 3 + 5$
${k}_{3} = 3 + 5 + 5$
${k}_{4} = 3 + 5 + 5 + 5$
...
${k}_{i} = 3 + \left(i - 1\right) \cdot 5$

Thus the sum can be written:
$\setminus {\Sigma}_{i = 1}^{66} \left[3 + \left(i - 1\right) \cdot 5\right]$
which can be simplified to:
$\setminus {\Sigma}_{i = 1}^{66} 3 + 5 \cdot \setminus {\Sigma}_{i = 1}^{66} \left[\left(i - 1\right)\right]$
and even to:
$\setminus {\Sigma}_{i = 1}^{66} 3 + 5 \cdot \left[\setminus {\Sigma}_{i = 1}^{66} i - \setminus {\Sigma}_{i = 1}^{66} 1\right]$
so:
$66 \cdot 3 + 5 \cdot \left[67 \cdot \frac{66}{2} - 66\right]$
$= 10923$

Dec 7, 2017

${S}_{66} = 10923$

Explanation:

$\text{the sum to n terms of an arithmetic sequence is}$

•color(white)(x)S_n=n/2[2a+(n-1)d]#

$\text{where a is the first term and d the common difference}$

$\text{here "a=3" and } d = 18 - 13 = 13 - 8 = 8 - 3 = 5$

$\Rightarrow {S}_{66} = 33 \left[\left(2 \times 3\right) + \left(65 \times 5\right)\right]$

$\textcolor{w h i t e}{\Rightarrow {S}_{66}} = 33 \left(6 + 325\right)$

$\textcolor{w h i t e}{\Rightarrow {S}_{66}} = 10923$