In how many ways

Find $1+(1+2)+(1+2+3)+\cdots +(1+2+3+\cdots+n)$ We write the above finite sum in terms of double dependent sum as

Solution 1 (Using standard formula) $F(n)=\sum_{k=1}^{n}\sum_{m=1}^{k} m =\frac{1}{2}\sum_{k=1}^{n}k(k+1)=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)=\frac{n+2}{3}T_n$

Solution 2 (Without using standard formula)

Now suppose we try to evaluate the sum without using the standard formula. To do so we note that for $k\leq n$ $\underbrace{1+1+\cdots +1}_{n}+\underbrace{2+2+\cdots +2}_{n-1} +\cdots +\underbrace{10+10+\cdots +10}_{n-9}+\cdots \underbrace{m+m+\cdots +m}_{n-m+1}$ So $F(n)=\sum_{k=1}^{n} \sum_{m=1}^{k}m = \sum_{k=1}^{n}\sum_{m=1}^{n-k+1} k=\sum_{k=1}^{n}\sum_{m=1}^{n} k -\sum_{k=1}^{n}\sum_{m=1}^{k}(m+1)+2\sum_{k=1}^n k$ $F(n)=\sum_{k=1}^{n+2}\sum_{k=1}^n k-2\sum_{k=1}^{n}\sum_{m=1}^k m \Rightarrow 3F(n)=(n+2)\sum_{k=1}^n k \Rightarrow F(n)=\frac{(n+2)}{3}\sum_{k=1}^n k$ $=\frac{n+2}{3}\sum_{k=1}^n\left(\frac{k+(n-k+1) }{2}\right) = \frac{n+2}{6}\sum_{k=1}^{n}(n+1)=\frac{n(n+1)(n+2)}{6}={n+2\choose 3}$

Solution 3 (Using combinatorial identities) $\sum_{k=1}^{n} \sum_{m=1}^{m}m =\sum_{k=1}^{n}T_k=\sum_{k=1}^{n}{ k+1\choose 2} =\frac{1}{4}{ 2n+2\choose 3} +{ n+1\choose 2} \overbrace{=}^{simplify} {n+2\choose 3}$ Solution 4 : One can try up with ordinary generating function.