3 times summation

Algebra Level 4

$\displaystyle \frac{1}{729}\times\sum_{a=1}^9\sum_{b=1}^9\sum_{c=1}^9(abc+ab+bc+ca+a+b+c) = \, ?$

The answer is 215.

1 solution

Nishant Sharma
Oct 17, 2014

Looks difficult at the first sight..Let's see if we can factorize..

Observe that $(a+1)(b+1)(c+1)=abc+ab+bc+ac+a+b+c+1$

$\implies\,abc+ab+bc+ac+a+b+c=(a+1)(b+1)(c+1)-1$

Now consider summation from right( as decided by the rule)..

$\sum\limits_{c=1}^9 (a+1)(b+1)(c+1)-1$

or $\sum\limits_{c=1}^9 (a+1)(b+1)c+\sum\limits_{c=1}^9 (a+1)(b+1)-\sum\limits_{c=1}^9 1$

which evaluates to $54(a+1)(b+1)-9$ . Doing this again for the subsequent summations we find that the given triple summation equals $54^3-9^3=9^3(6^3-1)=729\times215$ .

So our answer is $\boxed{215}$ .

$\sum_{a}\sum_{b}\sum_{c}(a + 1)(b+1)(c+1)$ can be written as $\sum_{a}(a+1)\sum_{b}(b+1)\sum_{c}(c+1)$ because the indices are independent of one another. Factoring further, it can be written as, $(\sum_{a}(a+1))(\sum_{b}(b+1))(\sum_{c}(c+1)) = 54 * 54 * 54$ Thereby reducing the calculation even further.

Avijit Sarker - 6 years, 7 months ago

How not understood

sandeep Rathod - 6 years, 7 months ago

Very nice problem! I used the same approach and the solution happens to be pretty easy. Thanks!

M Dub - 5 years, 10 months ago

That was fun. One can even generalize to a summation over $n$ independent indices running from $1$ to $n^2$ : $\frac{1}{n^{2n}}\sum_{\vec{\sigma}\in S_n(n^2)} \left(\prod_{k=1}^n(\sigma_k+1)-1\right) = \left(\frac{1}{2}(n^2+1)+1\right)^n-1,\qquad S_k(n):=\{1;\:\ldots;\:n\}^k$

Carsten Meyer - 1 month, 4 weeks ago

3 times summation

The answer is 215.

1 solution

0 pending reports