Let \(a = \max(m, n)\) and \(b = \min(m, n)\). Let \(T_n\) denote the \(n^{\text{th}}\) triangular number.

If \(b \geq \frac{a}{2}\), then number of triangles is equal to \(2 T_a + 2 T_b + 4 b\).
If \(b < \frac{a}{2}\), then number of triangles is equal to \(2 ( T_a - T_{a - 2b}) + 2 T_b + 4b\).
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Pranshu Gaba
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2 years, 2 months ago

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For n>m..and m>=ceiling function(n/2+0.5)..Generalization will be (n)(n+1)+(m)(m+5)..
For n>m.. and m<ceiling function(n/2+0.5)..and for n=odd.. (n)(n+1)+(m)(m+5)-Sum(k)..Where k=(ceiling function(n/2+0.5)-m)/0.5...And for n=Even.. k=((ceiling function(n/2+0.5)-m)/0.5-1)
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Charlz Charlizard
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2 years, 2 months ago

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TopNewestLet \(a = \max(m, n)\) and \(b = \min(m, n)\). Let \(T_n\) denote the \(n^{\text{th}}\) triangular number.

If \(b \geq \frac{a}{2}\), then number of triangles is equal to \(2 T_a + 2 T_b + 4 b\).

If \(b < \frac{a}{2}\), then number of triangles is equal to \(2 ( T_a - T_{a - 2b}) + 2 T_b + 4b\). – Pranshu Gaba · 2 years, 2 months ago

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For n>m..and m>=ceiling function(n/2+0.5)..Generalization will be (n)

(n+1)+(m)(m+5).. For n>m.. and m<ceiling function(n/2+0.5)..and for n=odd.. (n)(n+1)+(m)(m+5)-Sum(k)..Where k=(ceiling function(n/2+0.5)-m)/0.5...And for n=Even.. k=((ceiling function(n/2+0.5)-m)/0.5-1) – Charlz Charlizard · 2 years, 2 months agoLog in to reply