Power Sum

We can prove the more general result

$\displaystyle\sum_{\{a_{1},\dots,a_{k}\}\subseteq\{1,2,\dots,n\}}\frac{1}{a_{1}\cdots a_{k}}=n$

by mathematical induction.

It is easy to prove that the statement holds for n = 1.

Now suppose that it holds for n = p. On moving from p to (p + 1) we must consider all the subsets which do not include (p + 1) (let's call them the old subsets) plus the new subsets which will include the element (p + 1).

The old subsets are identical to those in the case when n = p. We know from the induction hypothesis that these old subsets contribute p to the new sum.

The new subsets include the set with the single element (p + 1), which contributes $\frac{1}{p+1}$ to the new sum.

The rest of the new subsets will consist of one of the old subsets with the addition of the element (p+1). Their total contribution to the new sum will be $\frac{1}{p+1}$ times the old sum (which is p by hypothesis)

Combining these three sorts of terms in the new sum we get the total sum for n = p+1 -

$p+\frac{1}{p+1}+\frac{p}{p+1}=p+\frac{p+1}{p+1}=p+1$ as we had hoped.

Noting that the hypothesis holds for n = 1, we can now invoke the principle of mathematical induction to prove the general case, and then the special result for n = 5 is rather easy!

$We\quad define\quad \sum { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 } } .....{ a }_{ k }\quad as\quad the\quad sum\quad of\quad \left\{ { a }_{ n } \right\} \quad taken\quad k\quad at\quad a\quad time.\quad \\ We\quad need\quad to\quad evaluate\quad S=\sum { \frac { 1 }{ { a }_{ 1 } } } +\sum { \frac { 1 }{ { a }_{ 1 }{ a }_{ 2 } } } +\sum { \frac { 1 }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 } } + } \sum { \frac { 1 }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 } } } +\frac { 1 }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 } } \quad ,\quad \left\{ { a }_{ n } \right\} =\left\{ 1,2,3,4,5 \right\} \\ Now\quad S=\quad \frac { \sum { { a }_{ 1 } } +\sum { { a }_{ 1 }{ a }_{ 2 } } +\sum { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 } } +\sum { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 } } +1 }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 } } =\quad \frac { (1+{ a }_{ 1 })(1+{ a }_{ 2 })(1+{ a }_{ 3 })(1+{ a }_{ 4 })(1+{ a }_{ 5 })-{ a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 } }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 } } =\quad \frac { 6!-5! }{ 5! } =\quad \boxed { 5 }$

This summation can be written as ...... [1+(1/1]×[1+(1/2)].............. ×[1+(1/5)]-1

Let $x$ be the value of the summation. Let $a_{n} = \sum\limits_{ S= { a_{1}, a_{2}, ..., a_{k}} \subseteq {1, 2, ..., 5}, |S|= n} a_{1}a_{2}...a_{k}$ .

It is easy to simplify that $x= \frac{a_{4}+ a_{3}+ a_{2}+ a_{1} + 1}{ a_{5}}$ with $a_{5} = 120$ .

It is easy to obtain $a_{4} + a_{3} + ...+ a_{1} + 1 = (1+1)(2+1)(3+1)(4+1)(5+1) - 5! = 600$ .

Thus we have $x = \frac{600}{120} =5$ .