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 solutionss.

Problems in Mathematics of KöMaL, November 2013