Let $1*S_n$ denote $1+2+3+4+5...+n.$
Let $2*S_n$ denote $1*S_1 + 1*S_2 + 1*S_3 +... + 1*S_n.$
Similarly, let $3*S_n$ denote $2*S_1 + 2*S_2 + 2*S_3 +... + 2*S_n.$
and so on.
Find the formula for $n*S_n.$
Note: $n*S$ is not multiplication. It is a new notation.
An example computation to help you find a general formula for $k\star S_n$ which may be proven by induction. Observe that $$ 3\star S_n=\sum_{m=1}^n2\star S_{m}=\sum_{m=1}^n\sum_{j=1}^m1\star S_j= \sum_{m=1}^n\sum_{j=1}^m\sum_{i=1}^j\binom{i}{1} $$ At this point use the identity $$ \sum_{i=0}^n\binom{i}{k}=\binom{n+1}{k+1} $$ repeatedly to unravel the sum. Indeed $$ \sum_{m=1}^n\sum_{j=1}^m\sum_{i=1}^j\binom{i}{1}= \sum_{m=1}^n\sum_{j=1}^m\binom{j+1}{2} =\sum_{m=1}^n\binom{m+2}{3} =\binom{n+3}{4}. $$ On the basis of this conmputation we may think that $$ k\star S_n=\binom{n+k}{k+1} $$ which you may prove by induction.