Question

the number of shots arranged in a complete pyramid the base of which is an equilateral triangle,each side containing n shots​

Answer 1

Answer:

Let's assume m and n be the number of short in the long and the short side respectively of the base

I want to figure out a equation which gives the number of shots present in the file in terms of m and n.

I tried doing this by considering the rows in AP. My answer is

n(n+1)(3m−n)6

This just a little bit off to the correct answer which is

n(n+1)(3m−n+1)6

Any suggestions on how to reach to the correct answer are welcome

Answer 2

Correct Question: Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.

The number of shots arranged in a complete pyramid the base of which is an equilateral triangle with each side containing n shots is equal to $\frac{n(n+1)(n+2)}{6}$

Step-by-step Explanation

Given: The base of the pyramid is an equilateral triangle and each side contains n shots.

To be found: To find the the number of shots arranged in a complete pyramid

Solution:

Given, a complete pyramid with the base as an  equilateral triangle and each side containing n shots.

So, the number of shots in the lowest or first layer of the pyramid can be given as,

$n + (n - 1) + (n - 2) +...+ 1= \frac{n(n + 1)/}{2} \\\implies n + (n - 1) + (n - 2) +...+ 1= \frac{n^{2} + n}{2}$

Similarly, using (n-1),(n-2),.., we can find the shots in the second, third, and other layers.

Hence, the total number of shots is calculated as,

$\frac{\sum(n^{2} + n)}{2} = \frac{{\sum n^{2} + \sum n}}{2} \\\implies \frac{\sum(n^{2} + n)}{2}= \frac{{[n(n+1)(2n+1)]/6 + [n(n+1)]/2}}{2}\\ \implies \frac{\sum(n^{2} + n)}{2}=\frac{ n(n+1)(n+2)}{6}$

Therefore, the total number of shots arranged in the complete pyramid is calculated as $\frac{n(n+1)(n+2)}{6}$

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