Summation of the fifth series is a compulsory course in mathematics of grade one in senior high school The general term formula of sequence {an} is an = (- 1) ^ (n-1) * (4n-3), and the sum of its first n terms can be obtained It must be adopted in time. Please use the multiply common ratio dislocation subtraction method

Summation of the fifth series is a compulsory course in mathematics of grade one in senior high school The general term formula of sequence {an} is an = (- 1) ^ (n-1) * (4n-3), and the sum of its first n terms can be obtained It must be adopted in time. Please use the multiply common ratio dislocation subtraction method

Your title is not very clear. I think it should be an = (4n-3) times (- 1) to the power of (n-1)
So we can clearly find a rule:
Let K be an integer and K ≥ 1
Then a2k + a2k-1 = - (8k-3) + (8k-4-3) = - 4, that is, the sum of even terms after odd term + is - 4
So a1 + A2 +. + a2k = - 4K
If n is even, let n = 2K, then a1 + A2 +. An = - 2n
When n is odd, let n = 2k-1, then a1 + A2 +. An-1 = - 2 (n-1) an = 4n-3 a1 + A2 +. An = - 2n + 2 + 4n-3 = 2N-1