Question #8d696

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Answer 1 · 2016-08-25T21:13:14

The general rule for any member ( $a_i$ ) of the given sequence we have:

$a_i=i^2+i+1$

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

Tony B

$color(brown)("Notice the next level of difference is 2")$
$4-2=2$
$6-4=2$
$8-6=2$
$10-8=2$

$color(blue)("So there is some form of progression involving 2")$

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

$a_1 = 1+2(1) =3$
$a_2=1+2(1+2)=7$
$a_3=1+2(1+2+3) = 13$
$a_4=1+2(1+2+3+4) = 21$
$a_5=1+2(1+2+3+4+5)=31$

Implying:

$a_i=1+2(1+2+...+i)$

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

$a_0=1+2(0)=1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Whenever you have a sequence of numbers like:

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

Look at:

$1+2+3 = 6 -> 3(1+3)/2$

$1+2+3+4=10=4(1+4)/2=10$

So we have [ count $xx$ mean value ]

So for any $a_i$ the count is $i$

Thus for any $a_i ->1+2[ i(1+i)/2]$

or if you prefer $a_i ->1+cancel(2)[(i+i^2)/(cancel(2))]$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Testing it!")$

$a_1->1+(1+1^2)=3$
$a_2->1+(2+2^2) = 7$
$a_3->1+(3+3^3)=13$
$a_4->1+(4+4^2)=21$
$a_5->1+(5+5^2)= 31$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("In conclusion:")$

The general rule for any member ( $a_i$ ) of the given sequence, we have:

$a_i=i^2+i+1$