Lessons from teaching gifted elementary students (Part 8f)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received, in the students’ original handwriting. They wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.


So far, I’ve used Pascal’s triangle to obtain

y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}

= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} +  \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}.

= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) +  \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right).

= \displaystyle \sum_{k=2}^{11}  \frac{11!}{(k-2)!(11-k)!}  +  3 \sum_{k=1}^{11}  \frac{11!}{(k-1)!(11-k)!}  + \sum_{k=0}^{11}  \frac{11!}{k!(11-k)!} .

= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!}  +  3 \sum_{j=0}^{10}  \frac{11!}{j!(10-j)!}  + \sum_{k=0}^{11}  \frac{11!}{k!(11-k)!} .

= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!}  +  3 \sum_{j=0}^{10}  11 \frac{10!}{j!(10-j)!}  + \sum_{k=0}^{11}  \frac{11!}{k!(11-k)!}

= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i}  +  33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11}  {11 \choose k}

To numerically evaluate y, I use the identity

\sum_{r=0}^n {n \choose r} = 2^n;

this identity can be proven by using the binomial theorem

\sum_{r=0}^n {n \choose r} x^r y^{n-r} = (x+y)^n

and then plugging in x = 1 and y = 1. Using this identity, I conclude that

y = 110 \times 2^9 + 33 \times 2^{10} + 2^{11}

= 55 \times 2 \times 2^9 + 33 \times 2^{10} + 2 \times 2^{10}

= 55 \times 2^{10} + 33 \times 2^{10} + 2 \times 2^{10}

= (55+33+2) \times 2^{10}

= 90 \times 2^{10}.

Since I know that 2^{10} = 1024, it’s now a simple matter of multiplication:

y = 90 \times 1024 = 92,160.

 (Trust me; after I showed my students this answer about five minutes after it was posed, I was ecstatic when I confirmed this answer with Mathematica.)

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