Discussion
What is the remainder when the sum \(1^3 + 2^3 + 3^3 + ... + 100^3\) is divided by 4?
Answer: A
The sum of the first n cubes is given by the formula \(S_n = [\frac{n(n+1)}{2}]^2\). Here, n=100. Sum = \([\frac{100(101)}{2}]^2 = (50 \times 101)^2 = (5050)^2\). We need to find the remainder of \(5050^2 \div 4\). First, find the remainder of \(5050 \div 4\). The number formed by the last two digits is 50. \(50 \div 4\) gives a remainder of 2. So, we need to find the remainder of \(2^2 \div 4 = 4 \div 4\). The remainder is 0.
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