A problem of summation of mathematical sequence in Senior High School Sum: SN = (1 / 1 × 3) + (4 / 3 × 5) + (9 / 5 × 7) + +Square of N / (2n-1) (2n + 1)

A problem of summation of mathematical sequence in Senior High School Sum: SN = (1 / 1 × 3) + (4 / 3 × 5) + (9 / 5 × 7) + +Square of N / (2n-1) (2n + 1)

No one answered after waiting for a long time. I'd better come
be careful
N ^ 2 / ((2n-1) (2n + 1)) = 1 / 4 + 1 / 8 * (1 / (2n-1) - 1 / (2n + 1)), each term is decomposed in this way, and then added together, the answer is (n ^ 2 + n) / (4N + 2)

2,2+4,2+4+6,…… ,2+4+6+…… +2n,…… And find s (n)

2,2+4,2+4+6,…… ,2+4+6+…… +2n,…… S (n) 1 + (1 + 2) + (1 + 2 + 3) +... + (1 + 2 + 3 +... N) = 1 * n + 2 * (n-1) + 3 * (n-2) +... + (n-1) * [n - (n-2)] + n * [n - (n-1)] = (n + 2n + 3N +... N * n) - [2 * (2-1) + 3 * (3-1) + 4 * (4-1) +... + n * (n-1)] = n * n * (n + 1) / 2 - (2 *)

Known arithmetic sequence AP: {an} = 2N-1
Known equal ratio sequence GP: {BN} = 2 ^ (n + 1)
Find t = A1B1 + a2b2 + a3b3 +. + anbn=_________
It is known that the general term formula of the arithmetic sequence an is 2N-1, and the general term formula of the arithmetic sequence BN is n + 1 power of 2. Find A1B1 plus a2b2 plus a3b3 plus. What is the value added to anbn?

Dislocation phase elimination method
Tn=1×2²+3×2³+ … +(2n-3) × 2 ^ n + (2n-1) × 2 ^ (n + 1) ① formula;
If both sides of the equation multiply by 2, 2tn = 1 × 2 & # 179; + 3 × 2 ^ 4 + +(2n-5) × 2 ^ n + (2n-3) × 2 ^ (n + 1) + (2n-1) × 2 ^ (n + 2) formula 2;
① - 2, we get - TN = 1 × 2 & # 178; + 2 × 2 & # 179; + 2 × 2 ^ 4 + +2×2^n+2×2^(n+1)－(2n-1)×2^(n+2)
=4+2^4+2^5+……… +2^(n+2)－(2n-1)×2^(n+2)
=4+2^(n+2)-2^4－(2n-1)×2^(n+2)
Then TN = (n-1) × 2 ^ (n + 3) + 12