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Problem C. 895. (April 2007)

C. 895. The series of diagrams shown is made up of more and more dark regular triangles. The series is continued to the n-th diagram according to the rule that can be observed. How many dark triangles are used altogether?

(5 pont)

Deadline expired on 15 May 2007.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A sötét háromszögek száma:

1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+n)=\frac{2\cdot1}{2}+
\frac{3\cdot2}{2}+\frac{4\cdot3}{2}+\cdots+\frac{(n+1)\cdot n}{2}=

=\frac{(1+1)1+(2+1)2+(3+1)3+\cdots+(n+1)n}{2}=\frac{(1^2+2^2+3^2+\cdots+n^2)+1\cdot(1+2+3+\cdots+n)}{2}=

=\frac{1}{2}\cdot\left(\frac{n(n+1)(2n+1)}{6}+\frac{(n+1)n}{2}\right)=\frac{n(n+1)(n+2)}{6}.


Statistics:

>
211 students sent a solution.
5 points:118 students.
4 points:8 students.
3 points:8 students.
2 points:16 students.
1 point:26 students.
0 point:10 students.
Unfair, not evaluated:25 solutions.

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