A sequence {an}: when n is odd, an = 5N + 1; when n is even, an = 2n2. Find the sum of the first 2m terms of the sequence (M is a positive integer)

A sequence {an}: when n is odd, an = 5N + 1; when n is even, an = 2n2. Find the sum of the first 2m terms of the sequence (M is a positive integer)

Because a2k + 1-a2k-1 = [5 (2k + 1) + 1] - [5 (2k-1) + 1] = 10, A1, A3, A5, a2m-1 are the arithmetic sequence with tolerance of 10. Because a2k + 2 △ a2k = (22K + 22) / (22k2) = 2, A2, A4, A6, A2M are the arithmetic sequence with common ratio of 2, so the sum of the first 2m terms of the sequence {an} is S2M = (a1 + a3 + A5 +) +a2m-1）+（a2+a4+a6+… +a2m）=[6+5(2m−1)+1]m2+2(1−2m)1−2=5m2+m+2m+1-2．