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Question Number 186631 by DAVONG last updated on 07/Feb/23

Commented by mr W last updated on 08/Feb/23

all answers given are wrong. a_k ∈N ⇒a_k ≥1 since 1^2 +2^2 +3^2 +...+50^2 =((50×51×101)/6)=42925 (a_1 )^2 +(2a_2 )^2 +(3a_3 )^2 +...(50a_(50) )^2 =42925 has only one solution: a_1 =a_2 =a_3 =...=a_(50) =1 so a_1 +a_2 +a_3 +...+a_(50) is and can only be equal to 50.

$${all}\:{answers}\:{given}\:{are}\:{wrong}. \\ $$$${a}_{{k}} \:\in{N}\:\Rightarrow{a}_{{k}} \geqslant\mathrm{1} \\ $$$${since} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{50}^{\mathrm{2}} =\frac{\mathrm{50}×\mathrm{51}×\mathrm{101}}{\mathrm{6}}=\mathrm{42925} \\ $$$$\left({a}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{2}{a}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{3}{a}_{\mathrm{3}} \right)^{\mathrm{2}} +...\left(\mathrm{50}{a}_{\mathrm{50}} \right)^{\mathrm{2}} =\mathrm{42925} \\ $$$${has}\:{only}\:{one}\:{solution}: \\ $$$${a}_{\mathrm{1}} ={a}_{\mathrm{2}} ={a}_{\mathrm{3}} =...={a}_{\mathrm{50}} =\mathrm{1} \\ $$$${so}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{\mathrm{50}} \:{is}\:{and}\:{can}\:{only}\: \\ $$$${be}\:{equal}\:{to}\:\mathrm{50}. \\ $$

Commented by mr W last updated on 09/Feb/23

correct question see Q186732

$${correct}\:{question}\:{see}\:{Q}\mathrm{186732} \\ $$

Answered by mahdipoor last updated on 07/Feb/23

B={a_1 ,a_2 ,...,a_(50) } , a_i ∈N f(B)=(a_1 )^2 +(2a_2 )^2 +...+(50a_(50) )^2 for B_1 ={1,1,...,1} , f(B_1 )=1^2 +2^2 +...+50^2 = (((50)×(50+1)×(2×50+1))/6)=42925 ⇒a_1 +a_2 +...+a_(50) =1+2+...+50=((50×51)/2)= 1275

$${B}=\left\{{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,...,{a}_{\mathrm{50}} \right\}\:\:\:,\:\:{a}_{{i}} \in{N} \\ $$$${f}\left({B}\right)=\left({a}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{2}{a}_{\mathrm{2}} \right)^{\mathrm{2}} +...+\left(\mathrm{50}{a}_{\mathrm{50}} \right)^{\mathrm{2}} \\ $$$${for}\:\:{B}_{\mathrm{1}} =\left\{\mathrm{1},\mathrm{1},...,\mathrm{1}\right\}\:\:\:,\:\:\:{f}\left({B}_{\mathrm{1}} \right)=\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +...+\mathrm{50}^{\mathrm{2}} = \\ $$$$\frac{\left(\mathrm{50}\right)×\left(\mathrm{50}+\mathrm{1}\right)×\left(\mathrm{2}×\mathrm{50}+\mathrm{1}\right)}{\mathrm{6}}=\mathrm{42925}\: \\ $$$$\Rightarrow{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{\mathrm{50}} =\mathrm{1}+\mathrm{2}+...+\mathrm{50}=\frac{\mathrm{50}×\mathrm{51}}{\mathrm{2}}= \\ $$$$\mathrm{1275} \\ $$

Commented by mr W last updated on 07/Feb/23

the question is strange. when a_1 =a_2 =a_3 =...=a_(50) =1 such that (1×1)^2 +(2×1)^2 +(3×1)^2 +...+(50×1)^2 =42925, then a_1 +a_2 +a_3 +...+a_(50) =50, not 1275. when a_1 =1, a_2 =2, ..., a_(50) =50 such that a_1 +a_2 +a_3 +...+a_(50) =((50×51)/2)=1275, then (a_1 )^2 +(2a_2 )^2 +...+(50a_(50) )^2 = 1^4 +2^4 +3^4 +...+50^4 =((50×51×101×7649)/(30))=65666765 ≠42925

$${the}\:{question}\:{is}\:{strange}. \\ $$$${when}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} ={a}_{\mathrm{3}} =...={a}_{\mathrm{50}} =\mathrm{1}\:{such}\:{that} \\ $$$$\left(\mathrm{1}×\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}×\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}×\mathrm{1}\right)^{\mathrm{2}} +...+\left(\mathrm{50}×\mathrm{1}\right)^{\mathrm{2}} =\mathrm{42925}, \\ $$$${then}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{\mathrm{50}} =\mathrm{50},\:{not} \\ $$$$\mathrm{1275}. \\ $$$${when}\:{a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{2},\:...,\:{a}_{\mathrm{50}} =\mathrm{50}\:{such}\:{that} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{\mathrm{50}} =\frac{\mathrm{50}×\mathrm{51}}{\mathrm{2}}=\mathrm{1275}, \\ $$$${then}\:\left({a}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{2}{a}_{\mathrm{2}} \right)^{\mathrm{2}} +...+\left(\mathrm{50}{a}_{\mathrm{50}} \right)^{\mathrm{2}} = \\ $$$$\mathrm{1}^{\mathrm{4}} +\mathrm{2}^{\mathrm{4}} +\mathrm{3}^{\mathrm{4}} +...+\mathrm{50}^{\mathrm{4}} \\ $$$$=\frac{\mathrm{50}×\mathrm{51}×\mathrm{101}×\mathrm{7649}}{\mathrm{30}}=\mathrm{65666765} \\ $$$$\neq\mathrm{42925} \\ $$

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