# Question #1d0d6

Jul 6, 2017

$\textcolor{red}{\text{Solution part 1 of 2}}$

Also see part 2 of 2 for the calculation method

If you use a strait line of best fit you can not be precise enough to satisfactorily predict values.

#### Explanation:

Triangular numbers are constructed as in the diagram.

You add up all the dots from the first one down to whichever point you wish to stop.

The ${11}^{\text{th}}$ term works out to be 66

Jul 6, 2017

$\textcolor{red}{\text{Solution part 2 of 2}}$ showing the quadratic

See part 1 of 2 first before reading this one.

#### Explanation:

The sequence is
$1 \text{ "3" "6" "10" "15" } 21. . .$

$1 \leftarrow \text{ 1st term}$

$1 + 2 = 3 \leftarrow \text{ 2nd term}$

$1 + 2 + 3 = 6 \leftarrow \text{ 3rd term}$

$1 + 2 + 3 + 4 = 10 \leftarrow \text{ 4th term}$

$1 + 2 + 3 + 4 + 5 = 15 \leftarrow \text{ 5th term}$

$1 + 2 + 3 + 4 + 5 + 6 = 21 \leftarrow \text{ 6th term}$

Notice that the last value in the sum is the term number or line number.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider any one of the rows

Let the first value be $f$ which in each case is 1

Let the last value be $L$

Then the sum of any row is $\text{count"xx"mean}$

count $= L$ so we have:

$\text{count"xx"mean"" "->" "(f+L)/2xxL" "=" } \frac{f}{2} L + \frac{1}{2} {L}^{2}$

but $f = 1$ giving:

$\frac{1}{2} L + \frac{1}{2} {L}^{2}$

Changing the order we have:

$\frac{1}{2} {L}^{2} + \frac{1}{2} L + 0$

Compare this to the standardised equation of a quadratic

$y = \frac{1}{2} {L}^{2} + \frac{1}{2} L + 0$
$y = a {x}^{2} \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} b x + c$

so if ${a}_{11}$ is the 11th term we have:

${a}_{11} = \frac{1}{2} \left({11}^{2}\right) + \frac{1}{2} \left(11\right)$

${a}_{11} = 60 \frac{1}{2} + 5 \frac{1}{2} = 66$