Cannon firing bullets into a log of mass 100m₀

[Biker]

The sum should indeed be from 2 to N since, i think, for example if i have 3 bullets then i have to add 2 intervals:
$$\sum_2^N \frac{N+99}{100}=\sum_2^N N+(N-1)99=\frac{(N-1)(N+2)}{2}+99(N-1)=\frac{(n-1)(N+200)}{200}$$
But it's strange that if i take the sum from 1 to (N-1) it gives a different thing:
$$\sum_1^{N-1} \frac{N+99}{100}=\frac{N(N+197)}{200}$$
I thought only the number of members in the sum counts, but no.

If you take the sum from 1 to N-1

with this function of impact time: $F(N) = \frac{N+99}{100} t_o$
You are adding up the value $t_o$ into account which is not asked in the question, Also you are neglecting the impact time between the Nth bullet and N-1th

 

[Karol]

Thanks Biker