## Tuesday, May 23, 2006

### Productive! Topic: Algebra/NT. Level: AIME.

Problem: (2006 Bellevue BATH Team) Evaluate $\displaystyle \sum_{a=1}^{10} \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} dcab$.

Solution: Note that we can write the sum as

$\displaystyle \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} (1 \cdot bcd+ 2 \cdot bcd + \cdots + 10 \cdot bcd) = \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} (1+2+\cdots+10)bcd$.

Using the same idea, we can write it as

$(1+2+\cdots+10)(1+2+\cdots+10)(1+2+\cdots+10)(1+2+\cdots+10) = 55^4$.

QED.

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Comment: Evaluating multiple summations this way is very effective. Other examples include factoring the harmonic series

$\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots = \left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right) \left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right) \left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right) = \prod \sum_{i=0}^\infty \frac{1}{p^i}$

or, more generally, any integer value of the Riemann Zeta Function

$\displaystyle \zeta(s) = 1^s+\frac{1}{2^s}+\frac{1}{3^s}+\cdots = \prod \sum_{i=0}^\infty \frac{1}{p^is}$.

Furthermore, since

$\displaystyle \sum_{i=0}^\infty \frac{1}{p^is}$

is an infinite geometric series of common ratio $\frac{1}{p^s}$, we can say that

$\zeta(s) = \prod \frac{1}{p^s-1}$,

where the product is taken over all primes $p$.