Problem

Find the sum of the series: \(\sum_{n=1}^{5} 2n^2 + 3n + 1\)

Answer

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Answer

Finally, we add these sums together to find the sum of the original series: \(110 + 45 + 5 = 160\).

Steps

Step 1 ：First, we break the series into three separate sums: \(\sum_{n=1}^{5} 2n^2\), \(\sum_{n=1}^{5} 3n\), and \(\sum_{n=1}^{5} 1\).

Step 2 ：We know that \(\sum_{n=1}^{k} n^2 = \frac{k(k + 1)(2k + 1)}{6}\), \(\sum_{n=1}^{k} n = \frac{k(k + 1)}{2}\), and \(\sum_{n=1}^{k} 1 = k\).

Step 3 ：So, \(\sum_{n=1}^{5} 2n^2 = 2\frac{5(5 + 1)(2\cdot5 + 1)}{6} = 110\), \(\sum_{n=1}^{5} 3n = 3\frac{5(5 + 1)}{2} = 45\), and \(\sum_{n=1}^{5} 1 = 5\).

Step 4 ：Finally, we add these sums together to find the sum of the original series: \(110 + 45 + 5 = 160\).

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