Find the $2011^{th}$ term of the sequence $2,3,5,6,7,8,10,11,…$
(a) $2056$
(b) $2011$
(c) $2013$
(d) $2060$
My Approach:
Let $2,3,5,6,7,8$ be one set of numbers.
In every 6 terms we reach 8 terms forward. From 1 to 9, 6 terms are present but the jump is of 8.
$2011/6=335+1/6$.
Hence 335 such whole sets and 1st term of the 336th set would give us the 2011th term.
335 full sets starting from 1, each of a jump 8 will reach $2+(335)(8)=2682$. Now the first term of the next set is $2683$.
But this doesn't match with any of the options given. Have I identified the sequence wrongly or is there a flaw in my logic?
Best Answer
If $a_n$ is the given sequence then $a_n-n=$ the number of squares less than or equal to $a_n$ which equals $\left\lfloor \sqrt{a_n}\right\rfloor $
Therefore we need to solve for $m$
$$m-\left\lfloor \sqrt{m}\right\rfloor =2011$$
This leads to the inequalities
$$2010<m-\sqrt{m}\leq 2011$$
and so
$$2055.34<m\leq 2056.35$$
and
$$m=2056$$