Problem C. 895.

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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 points)

Deadline expired.

Sorry, the solution is published in Hungarian only.

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 on problem C. 895.

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.

Problems in Mathematics of KöMaL, April 2007

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