Question

N, the set of natural numbers, is partitioned into subsets $S_1=\left\{1\right\},S_2=\{2,3\},S_3=\{4,5,6\},S_4=\{7,8,9,10\}$. The last term of these groups is $1,1+2,1+2+3,1+2+3+4$, so on. Find the sum of the elements in the subset $S_{50}$.

Answer 1

Last term of nth group$=1+2+3+...n$

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

The terms in nth group will be n and in A.P. of common difference $-1$ and first term being $\frac{n(n+1)}{2}$(i.e., last term as first term and $d=-1$ instead of $+1$)

$\therefore S_n=\frac{n}{2}[2\cdot \frac{n(n+1)}{2}+(n-1\left)\right(-1\left)\right]=\frac{n}{2}(n^2+1)$

$\therefore S_{50}=25\left(2501\right)=62525$.