# Essays/Pascal's Ladder

The numbers in Pascal's triangle satisfy, practically speaking, infinitely many identifies, so it's not too surprising that we can find some surprising relationships by looking closely.

— Graham, Knuth, & Patashni, *Concrete Mathematics*, page 155.

We consider the following sums of binomial coefficients, for` c <: d `:

+/ c ! d + i.n +/ (c + i.n) ! d + i.n

What are these sums? To use a particular example, the terms of the sums are

c=:2 [ d=:4 [ n=:5 c ! d + i.n 6 10 15 21 28 (c + i.n) ! d + i.n 6 10 15 21 28

and are adjacent entries in Pascal's triangle:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1

If we add the extra term` 4 `in Pascal's triangle to the right of` 6` ,
(or to the left of` 6 `in the mirror image of the above diagram reflected
about the vertical axis), the sum
collapses into a single binomial coefficient:

6 4 10 10 10 15 15 15 20 21 21 21 21 35 28 28 28 28 28 56 84 +/ 6 10 15 21 28 80 3!9 84 (3!9) - 3!4 80

which is` (1+c)!d+n`, ` `from which the original extra term` (1+c)!d `must
be subtracted. That is,

(+/ c ! d + i.n) = ((1+c)!d+n) - (1+c)!d (+/ (c+i.n) ! d+i.n) = ((c+n-1)!d+n) - (c-1)!d

"Pascal's ladder" or "Pascal's telescoping sum" would be good descriptions of this.

Of course, to make the derivation rigorous and respectable, one would have to employ a proof by (say) induction. The details of such a proof are left as an exercise for the reader.

### See also

- Hockey Stick Theorem or Christmas Stocking Theorem

Contributed by Roger Hui.