1.

There are 88 numbers a1,a2,a3,…,a88 and each of them is either equal to −3 or −1. Given that a21+a22+⋯+a288=280, then the value of a41+a42+⋯+a4884−500 is (correct answer + 3, wrong answer 0)

Answer»

There are 88 numbers a1,a2,a3,…,a88 and each of them is either equal to −3 or −1. Given that a21+a22+⋯+a288=280, then the value of a41+a42+⋯+a4884−500 is
(correct answer + 3, wrong answer 0)



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