Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1191. (November 2013)

C. 1191. A truckload of goods consists of n packages weighing 1 kg, n-1 packages weighing 2 kg, n-2 packages of 3 kg, and so on, finally there is 1 package of n kg. Express the average mass of a package as a function of n.

(5 pont)

Deadline expired on December 10, 2013.

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

Megoldás. Egy csomag átlagos tömege: $\displaystyle m=S/d$, ahol $\displaystyle S$ a csomagok tömegének összege, $\displaystyle d$ pedig a csomagok száma.

A csomagok száma: $\displaystyle d=n+(n-1)+(n-2)+...+2+1=\frac{(n+1)n}{2}$.

A csomagok tömegének összege:

$\displaystyle S=n\cdot1+(n-1)\cdot2+(n-2)\cdot3+...+(n+1-i)\cdot i+...+1\cdot n=$

$\displaystyle =((n+1)-1)\cdot1+((n+1)-2)\cdot2+((n+1)-3)\cdot3+...+((n+1)-i)\cdot i+...+((n+1)-n)\cdot n=$

$\displaystyle =((n+1)\cdot1-1\cdot1)+((n+1)\cdot2-2\cdot2)+((n+1)\cdot3-3\cdot3)+...+((n+1)\cdot i-i\cdot i)+ +...+((n+1)\cdot n-n\cdot n)=$

$\displaystyle =(n+1)(1+2+3+...+i+...+n)-(1^2+2^2+3^2+...+i^2+...n^2)=$

$\displaystyle =(n+1)\cdot\frac{(n+1)n}{2}-\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)n}{2}\cdot\left((n+1)-\frac{2n+1}{3}\right)=$

$\displaystyle =\frac{(n+1)n}{2}\cdot\frac{3n+3-2n-1}{3}=\frac{(n+1)n}{2}\cdot\frac{n+2}{3}.$

Tehát egy csomag átlagos tömege:

$\displaystyle m=\frac Sd=\frac{\frac{(n+1)n}{2}\cdot\frac{n+2}{3}}{\frac{(n+1)n}{2}}=\frac{n+2}{3}.$

### Statistics:

 170 students sent a solution. 5 points: 104 students. 4 points: 8 students. 3 points: 10 students. 2 points: 13 students. 1 point: 20 students. 0 point: 8 students. Unfair, not evaluated: 7 solutions.

Problems in Mathematics of KöMaL, November 2013