There are 88 numbers a1,a2,a3,…,a88 and each of them is either equal – InterviewSolution

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 $a_{1}, a_{2}, a_{3}, \dots, a_{88}$ and each of them is either equal to $- 3$ or $- 1.$ Given that $a_{1}^{2} + a_{2}^{2} + \dots + a_{88}^{2} = 280,$ then the value of $\frac{a_{1}^{4} + a_{2}^{4} + \dots + a_{88}^{4}}{4} - 500$ is (correct answer + 3, wrong answer 0)

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