Find the value of $(1*1)+(1*2*2)+(1*2*3*3)+$...$+(1*2$...$(n-1)*n*n)$.

Question

Find the value of $(1*1)+(1*2*2)+(1*2*3*3)+$...$+(1*2$...$(n-1)*n*n)$ .

What I Tried: I have absolutely no idea for this. I can write this as :- $$(1!*1) + (2!*2) + ... + (n!*n)$$

However, this can be simplified further. But I cannot understand how this can be simplified, except that I can write it as this :- $$\sum_{k=1}^{n} (k!*k)$$ Which seems probably stupid. The answer, however is given as $(n+1)! - 1$.

Can someone show and explain me how are we getting the answer?

Answer 1

Use $$(n+1)!-n!=n\cdot n!$$ and the telescopic summation. It gives your answer exactly.

Answer 2

Note that:

The $n^\text{th}$ term $T_n=k!k=k!(k+1-1)=k!(k+1)-k!$

This is a telescoping series;

put $k=1$, we get $T_1=2-1$

put $k=2$, we get $T_2=6-2$

put $k=3$, we get $T_3=24-6$

put $k=4$, we get $T_4=120-24$

and so on ...

put $k=n-1$,we get $n!-(n+1)!$

put $k=n$, we get $(n+1)!-n!$

When adding the terms, all the terms get cancelled except the second term in the first row and the first term in the last row, which gives us the sum as $(n+1)!-1$