# Question #8d696

Aug 25, 2016

The general rule for any member (${a}_{i}$) of the given sequence we have:

${a}_{i} = {i}^{2} + i + 1$

#### Explanation:

A trick always worth trying is to look for progressive increments.

$\textcolor{b r o w n}{\text{Notice the next level of difference is 2}}$
$4 - 2 = 2$
$6 - 4 = 2$
$8 - 6 = 2$
$10 - 8 = 2$

$\textcolor{b l u e}{\text{So there is some form of progression involving 2}}$

It took a bit of experimentation but I found the following progression:

${a}_{1} = 1 + 2 \left(1\right) = 3$
${a}_{2} = 1 + 2 \left(1 + 2\right) = 7$
${a}_{3} = 1 + 2 \left(1 + 2 + 3\right) = 13$
${a}_{4} = 1 + 2 \left(1 + 2 + 3 + 4\right) = 21$
${a}_{5} = 1 + 2 \left(1 + 2 + 3 + 4 + 5\right) = 31$

Implying:

${a}_{i} = 1 + 2 \left(1 + 2 + \ldots + i\right)$

To test this I took the sequence back further than the first given term to see if it worked. It did!

${a}_{0} = 1 + 2 \left(0\right) = 1$
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Whenever you have a sequence of numbers like:

$1 + 2 + 3 + 4 + . . + n$ there is a trick to finding the sum

Look at:

$1 + 2 + 3 = 6 \to 3 \frac{1 + 3}{2}$

$1 + 2 + 3 + 4 = 10 = 4 \frac{1 + 4}{2} = 10$

So we have [ count $\times$ mean value ]

So for any ${a}_{i}$ the count is $i$

Thus for any ${a}_{i} \to 1 + 2 \left[i \frac{1 + i}{2}\right]$

or if you prefer ${a}_{i} \to 1 + \cancel{2} \left[\frac{i + {i}^{2}}{\cancel{2}}\right]$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Testing it!}}$

${a}_{1} \to 1 + \left(1 + {1}^{2}\right) = 3$
${a}_{2} \to 1 + \left(2 + {2}^{2}\right) = 7$
${a}_{3} \to 1 + \left(3 + {3}^{3}\right) = 13$
${a}_{4} \to 1 + \left(4 + {4}^{2}\right) = 21$
${a}_{5} \to 1 + \left(5 + {5}^{2}\right) = 31$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{In conclusion:}}$

The general rule for any member (${a}_{i}$) of the given sequence, we have:

${a}_{i} = {i}^{2} + i + 1$