Question #b65ad

1 Answer
Aug 3, 2016

Answer:

The #n^"th"# term is given by #(n-1)(n+1)#

Explanation:

Examining the sequence, we find that the difference between successive terms increase by #2#, starting with an increase of #3#. That is, we have

#a_0 = 0#
#a_1 = 0 + 3#
#a_2 = 0 + 3 + 5#
#a_3 = 0 + 3 + 5 + 7#
#...#

As the difference increases by #2# each time, we can see that for each #k>=1#, the difference between #a_(k-1)# and #a_k# is #3 + 2(k-1)#. Thus, noting that the #n^"th"# term is #a_(n-1)# because we started from #a_0#, we have

#a_(n-1) = 0 + 3 + 5 + ... + (3+2((n-1)-1))#

#=(3+2(0))+(3+2(1))+...+(3+2(n-2))#

#=(3+3+...+3) + (2(0)+2(1) + ... + 2(n-2))#

#=3(n-1) + 2(1+2+...+(n-2))#

#=3(n-1)+2(((n-2)(n-1))/2)#

(The above comes from that the sum of all successive integers from #1# to #k# is given by #(k(k+1))/2#, with #k=n-2# in this case)

#=3(n-1)+(n-2)(n-1)#

#=(n-1)(n+1)#

Checking we find that this matches up with the given values. E.g. the #3^"rd"# term is #8 = 2*4 = (3-1)(3+1)#