Consider the sequence 1, 3, 3, 3, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, ... What is its 2016th term?

1 Answer
Feb 9, 2017

Answer:

#89#

Explanation:

Label the terms of the sequence #a_1, a_2, ..., a_2016, ...#.

Notice that the last term with the value #1# is #a_1#, the last term with the value #3# is #a_4#, the last term with the value #5# is #a_9#, and in general, the last term with the value #2k-1# is #a_(S_k)# where

#S_k = sum_(i=1)^k(2i-1) = 1+3+5+7+...+(2k-1)#

Then, to know the value of the terms in the subsequence containing #a_2016#, we need only find #k# such that #S_(k-1) < 2016 <= S_k#. Then we will have #a_2016 = 2k-1#. To do so, we will put #S_k# into a closed form.

#S_k = sum_(i=1)^k(2i-1)#

#=-k+2sum_(i=1)^ki#

#=-k+2((k(k+1))/2)#

#=-k+k(k+1)#

#=k^2#

(That the sum of the first #n# odd integers is #n^2# is a useful result to remember)

Examining squares, we find that

#S_44 = 44^2 < 2016 <= 45^2 = S_45#

Thus #a_2016# occurs after the final term of the subsequence consisting of #2(44)-1# but before the last term of the subsequence consisting of #2(45)-1#, giving us #a_2016 = 89#.