Question

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

Answer 1

$=1+2+3+...n=\frac{n(n+1)}{2}$ 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 intead of+1 ]

$S_2=\frac{n}{2}[2.\frac{n(n+1)}{2}+(n-1)(-1)]=\frac{n}{2}(n^2+1)$

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