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let $n>3$,if $a_{1}=a_{2}=\cdots=a_{n-1}=1$,and $a_{n}=-2$,find the following value $$f_{n}=\sum_{i=1}^{n}|a_{i}|-\sum_{1\le i<j\le n}|a_{i}+a_{j}|+\sum_{1\le i<j<k\le n}|a_{i}+a_{j}+a_{k}|-\cdots+(-1)^{n-1}|a_{1}+a_{2}+\cdots+a_{n}|$$

I have find the $n=4$,then $$f_{4}=5-6-4+3-1=-3$$ $$f_{5}=6-12-4+3\times 4-4-4+2=-4$$

math110
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1 Answers1

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First note the following identities

${n}\choose{1}$ - $2$${n}\choose{2}$ + $3$${n}\choose{3}$ - ... $=0$

${n}\choose{1}$ - ${n}\choose{2}$ + ${n}\choose{3}$ - ... $=1$.

The sums which do not include $a_n$ add up to

${n-1}\choose{1}$ - $2$${n-1}\choose{2}$ + $3$${n-1}\choose{3}$ - ... $=0$.

The sums which do include $a_n$ add up to

$2$${n-1}\choose{0}$ - $1$${n-1}\choose{1}$ + $0$${n-1}\choose{2}$ - $1$${n-1}\choose{3}$ + $2$${n-1}\choose{4}$- ...

$=2$ - ${n-1}\choose{1}$ + $2$${n-1}\choose{2}$ - $3$${n-1}\choose{3}$ + $4$${n-1}\choose{4}$- ... - $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...

$=2$ - $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...

$=2$ - $2$${n-1}\choose{1}$ + $2$${n-1}\choose{1}$- $2$${n-1}\choose{2}$ + $2$${n-1}\choose{3}$ - $2$${n-1}\choose{4}$- ...

$=2$ - $2$${n-1}\choose{1}$ + $2$

$=6-2n$

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