I think i got it ...

This is what given from the problem:

\(\frac{n(n+1)(n+2)}{6}\) = 8436

==> n(n+1)(n+2) = 8436 * 6

==> n(n+1)(n+2) = 2 * 2 * 3 * 19 * 37 * 6

==> n(n+1)(n+2) = (2 * 3 * 6) * 37 * (19 * 2)

==> n(n+1)(n+2) = 36 * 37 * 38

So, Ans: 36

By the way, What is radius information doing in this problem?

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Here is the tedious approach i followed to get \(\frac{n(n+1)(n+2)}{6}\)

Given:

1 + 3 + 6 + 10 + ... nth layer count= 8436

==>

(1) + (1 + 2) + (1 + 2 + 3) + (1+2+3+4) + ... nth layer count= 8436

==> Sum of first 1 numbers + Sum of first 2 numbers + Sum of first 3 numbers + Sum of first 4 numbers + ... + Sum of first (n-1) numbers + Sum of first n numbers = 8436

==> 1 + 3 + 6 + 10 + ... \(\frac{(n-1)n}{2}\) + \(\frac{n(n+1)}{2}\) = 8436

Let me reverse the above LHS as below for better reading ...

==> \(\frac{n(n+1)}{2}\) + \(\frac{(n-1)n}{2}\) + \(\frac{(n-2) (n-1)}{2}\) + \(\frac{(n-3) (n-2)}{2}\) + ... = 8436

==> Now take out the common denominator, then

\(\frac{n(n+1) + (n-1)(n) + (n-2)(n-1) + (n-3)(n-2) + ....}{2}{\) = 8436

==> ((\(n^2\) + n) + (\(n^2\) - n) + (\(n^2\) -3n +2) + (\(n^2\) -5n + 6) + ....) * (1/2) = 8436

Now add 2 successive terms together ...

i.e, (\(n^2\) + n) + (\(n^2\) - n) = \(2n^2\),

(\(n^2\) -3n +2) + (\(n^2\) -5n + 6) = 2\(n^2\) -8n +8

==> ( 2\(n^2\) + 2\(n^2\) - 8n +8 + ...) * (1/2) = 8436

Now, cancel out 2 from both numerator and denominator

==> \(n^2\) + (\(n^2\) - 4n + 4) + ... = 8436

==> \(n^2\) + \((n-1)^2\) + ... = 8436

Bascially this is what's given in the question:

\(n^2\) + \((n-1)^2\) + \((n-4)^2\) + \((n-6)^2\) + ... = 8436 ---- (*)

To test the above equation, the sum of first 3 layer counts is:

\(3^2\) + \((3-2)^2\) = 9 + 1 = 10, which is equal to the given numbers 1 + 3 + 6

Now, what (*) formulae represents is the sum of squares of even numbers <= n (where n = last layer)

==> (Sum of squares of first n numbers) - (Sum of squares of odd numbers <= n)

Sum of squares of first n numbers is: (\(\frac{n(n+1)(2n+1)}{6}\))

Sum of squares of first n odd numbers is: (n)(2n-1)(2n+1)/3. To find out sum of first (n/2) odd numbers substitue n = n/2 in this formulae.

==> (\(\frac{n(n+1)(2n+1)}{6}\)) - ((n/2)*(2*(n/2) - 1)*(2 * (n/2) + 1))/3

==> \(\frac{n(n+1)(n+2)}{6}\)

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Cheers!

Ravi

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Cheers!

Ravi

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