Question

Consider the sequence 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17 … of all positive integers that are not perfect squares. Determine the ${{2011}^{th}}$ term of this sequence.
(a) 2056
(b) 2011
(c) 2053
(d) 2055

Answer 1

Hint: To solve this problem we should first observe the given series carefully and we can see that numbers which are perfect squares of the natural numbers are missing from the series. So to find the ${{2011}^{th}}$ term of the series we have to know that how many terms are missing from before the 2011 for that we will find the number whose square will exceed the number 2011 and then add that number minus 1 with 2011 and see what number we get, if the obtained number contains any new perfect square we will subtract that from series and add 1 to that number to get the answer.

Complete step by step answer:
We are given the series 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17 …,
Now if we observe the above series carefully we can see that numbers with perfect squares are missing from the series like ${{1}^{2}}=1,\,{{2}^{2}}=4,\,{{3}^{2}}=9\,,..$ and so on.
And we have to find the ${{2011}^{th}}$ term of the series,
So to find the ${{2011}^{th}}$ term of this series we will have to add the number of perfect squares that come before 2011 with the number 2011,
Then if there is anymore extra perfect square that comes before the obtained number then we will have to add that number of perfect squares also,
So we perform the above methods as,
We know that ${{44}^{2}}=1936$ and ${{45}^{2}}=2025$,
So the number of perfect squares that comes before the 2011 are 44 as square of 45 is greater than 2011 and square of 44 is less than 2011,
Now if we add 44 with 2011, we get
44 + 2011 = 2055
Now if we see square of 45 i.e. ${{45}^{2}}=2025$ is also included inside the number 2055 but it is not in the series so 2055 is the ${{2010}^{th}}$ term of the series,
And if we add 1 with 2055 we will get the ${{2011}^{th}}$ term of the series, i.e.
1 + 2055 = 2056
So ${{2011}^{th}}$ term of the series is 2056.

So, the correct answer is “Option A”.

Note: Many students here can make mistake that they may forget to add 1 to the number 2055 to get the ${{2011}^{th}}$ term and then they may get the wrong answer as 2055 is also in the given options so you need to be careful about that. And also you can easily solve this problem by checking the options given, first subtract the 2011 from the given option then square the obtained number and if it is less than the option and its next term’s square is larger than the chosen option then it will be the correct answer.