Question

Question #8d696

Answer 1

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.

`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`