35) 등차수열 $\left\{a_{n}\right\}$ 의 첫째항부터 제 $n$ 항까지의 합을 $S_{n}$ 이 라 하면 $S_{n}=3 n^{2}-2 n$ 이다. 이때, $a_{2}+a_{4}+a_{6}+\cdots+a_{2 n}$ 을 구하시오.

Step 1 ：\(S_n = 3n^2 - 2n\)

Step 2 ：\(a_n = S_n - S_{n-1} = (3n^2 - 2n) - (3(n-1)^2 - 2(n-1))\)

Step 3 ：\(a_n = 3n^2 - 2n - 3(n^2 - 2n + 1) + 2n - 2\)

Step 4 ：\(a_n = 3n^2 - 2n - 3n^2 + 6n - 3 + 2n - 2\)

Step 5 ：\(a_n = -n^2 + 6n - 5\)

Step 6 ：\(a_{2n} = -(2n)^2 + 6(2n) - 5\)

Step 7 ：\(a_{2n} = -4n^2 + 12n - 5\)

Step 8 ：\(\sum_{k=1}^{n} a_{2k} = a_2 + a_4 + a_6 + \dots + a_{2n}\)

Step 9 ：\(\sum_{k=1}^{n} a_{2k} = \sum_{k=1}^{n} (-4k^2 + 12k - 5)\)

Step 10 ：\(\sum_{k=1}^{n} a_{2k} = -4\sum_{k=1}^{n} k^2 + 12\sum_{k=1}^{n} k - 5n\)

Step 11 ：\(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)

Step 12 ：\(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)

Step 13 ：\(\sum_{k=1}^{n} a_{2k} = -4\left(\frac{n(n+1)(2n+1)}{6}\right) + 12\left(\frac{n(n+1)}{2}\right) - 5n\)

Step 14 ：\(\sum_{k=1}^{n} a_{2k} = -\frac{4}{3}n(n+1)(2n+1) + 6n(n+1) - 5n\)

Step 15 ：\(\sum_{k=1}^{n} a_{2k} = n(n+1)\left(-\frac{4}{3}(2n+1) + 6 - \frac{5}{n+1}\right)\)

Step 16 ：\(\sum_{k=1}^{n} a_{2k} = n(n+1)\left(-\frac{8}{3}n + \frac{13}{3} - \frac{5}{n+1}\right)\)

Step 17 ：\(\boxed{\sum_{k=1}^{n} a_{2k} = n(n+1)\left(-\frac{8}{3}n + \frac{13}{3} - \frac{5}{n+1}\right)}\)