Menu Close

a-b-c-1-2-3-n-find-a-b-c-abc-




Question Number 87313 by mr W last updated on 03/Apr/20
a,b,c=1,2,3,...,n  find Σ_(a≠b≠c) abc
$${a},{b},{c}=\mathrm{1},\mathrm{2},\mathrm{3},…,{n} \\ $$$${find}\:\underset{{a}\neq{b}\neq{c}} {\sum}{abc} \\ $$
Commented by mr W last updated on 04/Apr/20
yes, you are right.
$${yes},\:{you}\:{are}\:{right}. \\ $$
Commented by mr W last updated on 04/Apr/20
for example n=4  Σ_(a≠b≠c) abc=1×2×3+1×2×4+1×3×4+2×3×4=50
$${for}\:{example}\:{n}=\mathrm{4} \\ $$$$\underset{{a}\neq{b}\neq{c}} {\sum}{abc}=\mathrm{1}×\mathrm{2}×\mathrm{3}+\mathrm{1}×\mathrm{2}×\mathrm{4}+\mathrm{1}×\mathrm{3}×\mathrm{4}+\mathrm{2}×\mathrm{3}×\mathrm{4}=\mathrm{50} \\ $$
Commented by mr W last updated on 03/Apr/20
to choose three different numbers  from n numbers there are C_3 ^n    combinations. from each combination  we get the product of the three numbers.  now it is to find the sum of all these  products.
$${to}\:{choose}\:{three}\:{different}\:{numbers} \\ $$$${from}\:{n}\:{numbers}\:{there}\:{are}\:{C}_{\mathrm{3}} ^{{n}} \: \\ $$$${combinations}.\:{from}\:{each}\:{combination} \\ $$$${we}\:{get}\:{the}\:{product}\:{of}\:{the}\:{three}\:{numbers}. \\ $$$${now}\:{it}\:{is}\:{to}\:{find}\:{the}\:{sum}\:{of}\:{all}\:{these} \\ $$$${products}. \\ $$
Commented by MJS last updated on 04/Apr/20
sorry but for n=4 we have  1×2×3+1×2×4+1×3×4+2×3×4=50
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{for}\:{n}=\mathrm{4}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{1}×\mathrm{2}×\mathrm{3}+\mathrm{1}×\mathrm{2}×\mathrm{4}+\mathrm{1}×\mathrm{3}×\mathrm{4}+\mathrm{2}×\mathrm{3}×\mathrm{4}=\mathrm{50} \\ $$
Answered by mr W last updated on 04/Apr/20
S=Σabc=(Σ_(a=1) ^n a)(Σ_(b=1) ^n b)(Σ_(c=1) ^n c)=[((n(n+1))/2)]^3     S_1 =Σ_(a=b=c) abc=Σ_(a=1) ^n a^3 =((n^2 (n+1)^2 )/4)    S_2 =Σ_(a,b=c≠a) abc=3Σ_(b=1) ^n [(Σ_(a=1) ^n a−b)(b^2 )]  =3Σ_(b=1) ^n [(Σ_(a=1) ^n a)b^2 −b^3 ]  =3[(Σ_(a=1) ^n a)(Σ_(b=1) ^n b^2 )−(Σ_(b=1) ^n b^3 )]  =3[((n(n+1))/2)×((n(n+1)(2n+1))/6)−((n^2 (n+1)^2 )/4)]  =(((n−1)n^2 (n+1)^2 )/2)    S_3 =Σ_(a≠b≠c) abc=(1/(3!))(S−S_1 −S_2 )  =(1/6)[((n^3 (n+1)^3 )/8)−((n^2 (n+1)^2 )/4)−(((n−1)n^2 (n+1)^2 )/2)]  =(((n−2)(n−1)n^2 (n+1)^2 )/(48))    example: n=4  S_3 =((2×3×4^2 ×5^2 )/(48))=50  1×2×3+1×2×4+1×3×4+2×3×3  =50    example: n=5  S_3 =((3×4×5^2 ×6^2 )/(48))=225  1×2×3+1×2×4+1×2×5+1×3×4+1×3×5+1×4×5  +2×3×4+2×3×5  +3×4×5  =225
$${S}=\Sigma{abc}=\left(\underset{{a}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\right)\left(\underset{{b}=\mathrm{1}} {\overset{{n}} {\sum}}{b}\right)\left(\underset{{c}=\mathrm{1}} {\overset{{n}} {\sum}}{c}\right)=\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{3}} \\ $$$$ \\ $$$${S}_{\mathrm{1}} =\underset{{a}={b}={c}} {\sum}{abc}=\underset{{a}=\mathrm{1}} {\overset{{n}} {\sum}}{a}^{\mathrm{3}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$${S}_{\mathrm{2}} =\underset{{a},{b}={c}\neq{a}} {\sum}{abc}=\mathrm{3}\underset{{b}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left(\underset{{a}=\mathrm{1}} {\overset{{n}} {\sum}}{a}−{b}\right)\left({b}^{\mathrm{2}} \right)\right] \\ $$$$=\mathrm{3}\underset{{b}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left(\underset{{a}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\right){b}^{\mathrm{2}} −{b}^{\mathrm{3}} \right] \\ $$$$=\mathrm{3}\left[\left(\underset{{a}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\right)\left(\underset{{b}=\mathrm{1}} {\overset{{n}} {\sum}}{b}^{\mathrm{2}} \right)−\left(\underset{{b}=\mathrm{1}} {\overset{{n}} {\sum}}{b}^{\mathrm{3}} \right)\right] \\ $$$$=\mathrm{3}\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$$=\frac{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${S}_{\mathrm{3}} =\underset{{a}\neq{b}\neq{c}} {\sum}{abc}=\frac{\mathrm{1}}{\mathrm{3}!}\left({S}−{S}_{\mathrm{1}} −{S}_{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{8}}−\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$$=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{48}} \\ $$$$ \\ $$$${example}:\:{n}=\mathrm{4} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{2}×\mathrm{3}×\mathrm{4}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} }{\mathrm{48}}=\mathrm{50} \\ $$$$\mathrm{1}×\mathrm{2}×\mathrm{3}+\mathrm{1}×\mathrm{2}×\mathrm{4}+\mathrm{1}×\mathrm{3}×\mathrm{4}+\mathrm{2}×\mathrm{3}×\mathrm{3} \\ $$$$=\mathrm{50} \\ $$$$ \\ $$$${example}:\:{n}=\mathrm{5} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{3}×\mathrm{4}×\mathrm{5}^{\mathrm{2}} ×\mathrm{6}^{\mathrm{2}} }{\mathrm{48}}=\mathrm{225} \\ $$$$\mathrm{1}×\mathrm{2}×\mathrm{3}+\mathrm{1}×\mathrm{2}×\mathrm{4}+\mathrm{1}×\mathrm{2}×\mathrm{5}+\mathrm{1}×\mathrm{3}×\mathrm{4}+\mathrm{1}×\mathrm{3}×\mathrm{5}+\mathrm{1}×\mathrm{4}×\mathrm{5} \\ $$$$+\mathrm{2}×\mathrm{3}×\mathrm{4}+\mathrm{2}×\mathrm{3}×\mathrm{5} \\ $$$$+\mathrm{3}×\mathrm{4}×\mathrm{5} \\ $$$$=\mathrm{225} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *