Tennis Ball Pyramid

The $n$ th layer contains the amount of tennis balls equal to the $n$ th triangle number, $\frac{n(n+1)}{2}$ . So, essentially, we're finding $\displaystyle\sum_{i=1}^{100}\frac{i(i+1)}{2}$ $=\displaystyle\frac{1}{2}\sum_{i=1}^{100}(i^2+i)$ $=\displaystyle\frac{1}{2}\left(\sum_{i=1}^{100}i^2+\sum_{i=1}^{100}i\right)$

Now, $\sum_{i=1}^ki=\frac{k(k+1)}{2}$ and $\sum_{i=1}^ki^2=\frac{k(k+1)(2k+1)}{6}$ , so the above expression equals $=\frac{1}{2}\left(\frac{100(100+1)(2\cdot100+1)}{6}+\frac{100(100+1)}{2}\right)$ $=171700$

If you number the layers of the pyramid from top to bottom, the $l$ -th layer consists of a triangle with the edge length $l$ . The triangle consists of $l$ rows with a total number of $m$ balls in each row, which ranges from 1 to $l$ . The total number of balls thus results from the sum of all layers and rows $N(n) = \sum_{l = 1}^n \sum_{m = 1}^{l} m$ In the case $n \ggg 1$ we can approximate the double sum by an double integral: $N(n) \approx \int_{0}^n \left(\int_{0}^{l} dm \right) dl = \frac{n^3}{6}$ In general, the function $N(n)$ is a polynomial with degree three: $N(n) = \frac{n^3}{6} + b \cdot n^2 + c \cdot n$ The coefficient of the $n^3$ term could already be determined by the integral estimation. The coefficients $b$ and $c$ can be determined by cases $n = 1$ and $n = 2$ : $\begin{aligned} & & N(1) = \frac{1}{6} + b + c &= 1 \\ & & N(2) = \frac{4}{3} + 4b + 2c &= 4 \\ \Rightarrow & & \left(\begin{array}{cc} 1 & 1 \\ 4 & 2 \end{array} \right) \cdot \left(\begin{array}{c} b \\ c \end{array} \right) &= \left(\begin{array}{c} ^5/_6 \\ ^8/_3 \end{array} \right) \\ \Rightarrow & & \left(\begin{array}{c} b \\ c \end{array} \right) &= -\frac{1}{2} \left(\begin{array}{cc} 2 & -1 \\ -4 & 1 \end{array} \right) \cdot \left(\begin{array}{c} ^5/_6 \\ ^8/_3 \end{array} \right) = \left(\begin{array}{c} ^1/_2 \\ ^1/_3 \end{array} \right)\\ \Rightarrow & & \boxed{N(n) = \dfrac{n^3}{6} + \dfrac{n^2}{2} + \dfrac{n}{3}} \end{aligned}$ That this formula holds for all $n \in \mathbb{N}$ , can be proved by complete induction. Suppose we have proved the validity of the formula for all numbers ranging from 1 to $n$ . Now we calculate $N (n + 1)$ : $\begin{aligned} N(n+1) &= N(n) + \sum_{m = 1}^{n + 1} m \\ &= \left(\frac{n^3}{6} + \frac{n^2}{2} + \frac{n}{3} \right) + \left(\frac{(n+2)(n + 1)}{2} \right) \\ &= \frac{n^3}{6} + n^2+ \frac{11}{6} n + 1 \\ &= \frac{n^3 + 3n^2 + 3n + 1}{6}+ \frac{n^2 + 2n + 1}{2} + \frac{n+1}{3} \\ &= \frac{(n+1)^3}{6} + \frac{(n+1)^2}{2} + \frac{(n+1)}{3} & \square \end{aligned}$ Thereby, we used the sum formulo of Gauss: $\sum_{m = 1}^{n} m = \frac{1}{2} (n+ 1) n$