Super Sigma

Algebra Level pending

Given that ∑ i = 1 50 a i 2 + 42925 = 2 ∑ i = 1 50 i a i \large \sum_{i=1}^{50} a_{i}^2 + 42925 = 2 \sum_{i=1}^{50} ia_{i} for a 1 , a 2 , … , a 50 a_{1}, a_{2}, \ldots , a_{50} are real numbers. Find ∑ i = 1 50 a i . \sum_{i=1}^{50} a_{i}.


The answer is 1275.

Skye Rzym by SKYE RZYM , 20, Indonesia

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1 solution

What we have to see is 1 2 + 2 2 + 3 2 + ⋯ + 5 0 2 = 42925 1^2+2^2+3^2+\cdots+50^2=42925 . After that it can be easily seen that, ( a 1 − 1 ) 2 + ( a 2 − 2 ) 2 + ( a 3 − 3 ) 2 + ⋯ + ( a 50 − 50 ) 2 = 0 ( a 1 − 1 ) = ( a 2 − 2 ) = ( a 3 − 3 ) = ⋯ = ( a 50 − 50 ) = 0 \begin{aligned}&(a_{1}-1)^2+(a_{2}-2)^2+(a_{3}-3)^2+\cdots+(a_{50}-50)^2=0\\&(a_{1}-1)=(a_{2}-2)=(a_{3}-3)=\cdots=(a_{50}-50)=0\end{aligned} ∑ i = 1 50 a i = 1 + 2 + 3 + ⋯ + 50 = 1275 \sum_{i=1}^{50} a_{i}=1+2+3+\cdots+50=\boxed{1275}

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