Question

Find the number of shots arranged in a complete pyramid having a rectangular base with m
and n number of shots in the long and short side respectively of the base.

please provide me with proper answer

No rubbish talks please​

Answer 1

Answer:

(1/6)×n(n+1)(3m−n+1)

Step-by-step explanation:

The base has mn shots, the next rectangle up has (m−1)(n−1) and so on up to the top "rectangle" which has (m−n+1)1.

We sum first the terms involving m:

m(n+(n−1)+ ⋯ +1)=(1/2) × mn(n+1)

We now have to subtract

1(n−1)+2(n−2)+3(n−3)+ ⋯ +(n−1)1

We can split that into

(1+2+⋯+(n−1))n − (1^2+2^2+⋯+(n−1)√2)

=1/2 × (n−1)n^2−1/6 × (n−1)n(2n−1)=1/6 ×n(n−1)(3n−2n+1)

So the final answer is

(1/6)×n(n+1)(3m−n+1)

EZ :)