Question #8d696

1 Answer
Aug 25, 2016

Answer:

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.

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#