Find the sum of the first 100 terms of the sequence defined by (a_n = n^2).

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Step:1 ：The sum of the first n terms of a sequence is given by the formula (S_n = sum_{i=1}^{n} a_i), where (a_i) is the ith term of the sequence. In this case, (a_i = i^2), so the sum of the first n terms is (S_n = sum_{i=1}^{n} i^2).Step:2 ：We can use the formula for the sum of the first n squares, which is (frac{n(n + 1)(2n + 1)}{6}). Plugging in n = 100, we get (S_{100} = frac{100(101)(201)}{6}).Step:3 ：Calculating the above expression gives us (S_{100} = 338350).

Answer

Step: 1 ：The sum of the first n terms of a sequence is given by the formula (S_n = sum_{i=1}^{n} a_i), where (a_i) is the ith term of the sequence. In this case, (a_i = i^2), so the sum of the first n terms is (S_n = sum_{i=1}^{n} i^2).Step: 2 ：We can use the formula for the sum of the first n squares, which is (frac{n(n + 1)(2n + 1)}{6}). Plugging in n = 100, we get (S_{100} = frac{100(101)(201)}{6}).Step: 3 ：Calculating the above expression gives us (S_{100} = 338350).

Find the sum of the first 100 terms of the sequence defined by \(a_n = n^2\).

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