Hallucinating Terms

The main problem with this sum is as we plunge ourselves more into the sum, we realize its a little bit difficult than we think.

Lets suppose that the $n^{th}$ term falls in the $k^{th}$ group that consists of k terms. ( Here groups mean the groups with same members, like (3,3,3). )

Thus the 1st term of the $k^{th}$ group is: The number of terms before it is $1+2+3........(k-1)=\frac{k(k-1)}{2}$ and so the first term is $\frac{k(k-1)}{2}+1$ . Similarly the last term of the $k^{th}$ group is $\frac{k(k+1)}{2}$ .

Had the $n^{th}$ term been the first term of any $k^{th}$ group, then

$\frac{k(k-1)}{2}+1=n$

$k^{2}-k+2-2n=0$

$k=\frac{1\pm\sqrt{1-4.1.(2-2n)}}{2}=\frac{1\pm\sqrt{8n-7}}{2}$

But this would have been the expression had the $n^{th}$ term been the first term of the $k^{th}$ group. In such cases both n and k would have been integers. We are left with one objective:

1) If n is increased then we should get k as output.

Upon thinking, we realize that there is another value of n , let n' to be such that it yields new value of k = k +1, i.e, at this value of n', the next group commences. For all values of n between n and n' , we should get k always. We get another clue: k is increasing with n , not linearly though. Thus before attaining the value k +1, the output would be k+ a fractional part.

Thus we use our Greatest Integer Function- Lo and behold! The expression is thus derived!

The value of the $n^{th}$ term is given by k as $k=floor(\frac{1+\sqrt{8n-7}}{2})$ . ( We neglect the negative sign, though, its not needed.)

Thus we are left with putting n =4100, thus k comes out to be $\boxed{91}$ .

Hope this helps. :)