I would like to ask you ∞Σ n = 0 (- 1) ^ n * (n ^ 2-N + 1) / 2 ^ n sum and ∞Σn = 1 2 ^ n / (3 ^ n (2n-1)) sum

I would like to ask you ∞Σ n = 0 (- 1) ^ n * (n ^ 2-N + 1) / 2 ^ n sum and ∞Σn = 1 2 ^ n / (3 ^ n (2n-1)) sum

Do it with power series;
This is the result (n = 1 to the infinite (n-1) of the sum (n = 1 to the infinite) (-1) ^ nNX (n-1) (n = 1 to the infinite (n = 1 to the infinite) (-1) (n = 1 to the infinite) (-1) ^ n (n = 1 to the infinite (n = 1 to the infinite) ((n = 1) ^ nNX (n-1) (n = 1 to the infinite (n = 1 to the infinite) (n = 1 to the infinite) (-1) ^ n (n (n (n (n (n (n))) ^ n (n (x ^ n (x ^ n ^ n ^ n (n ^ n))) \\\\\\\\\\\\\\\\\\ / (1 + 0.5)
The second problem is similar: F (x) = sum (n from 1 to infinity) 2 ^ NX ^ (2n-1) / 3 ^ n (2n-1), f '(x) = sum (n from 1 to infinity) 2 ^ NX ^ (2n-2) / 3 ^ n = 1 / x ^ 2 sum (n from 1 to infinity) 2 ^ NX ^ (2n) / 3 ^ n = 1 / x ^ 2 [(2 / 3x ^ 2). 1 - (2 / 3x ^ 2)] = 2 / (3-2x ^ 2), the original number = f (1) = 2

（2-3×5-1）+（4-3×5-2）+… （2n-3×5-n）=______ ．

The original formula = (2 + 4 +...) +2n）-3×（5-1+5-2+… +5-N) = n (2 + 2n) 2-3 × 5 − 1 (1 − 5 − n) 1 − 5 − 1 = n (n + 1) - 3 × 1 − 5 − N4 = n (n + 1) - 34 [1 - (15) n]

Sum: (A-1) + (a-178; - 2) + +(a∧n-n),(a≠0)

Wait a minute