Question

求数列{n^2*2^n}的前n项和（高二数列问题）

Answer 1

a(n) = n^2*2^n = n(n+1)2^n - n2^n = b(n) - c(n).

C(n) = c(1)+c(2)+c(3) + ... + c(n-1) + c(n)

= 1*2 + 2*2^2 + 3*2^3 + ... + (n-1)2^(n-1) + n2^n

2C(n) = 1*2^2 + 2*2^3 + ... + (n-1)2^n + n2^(n+1),

C(n) = 2C(n) - C(n) = - 2 - 2^2 - 2^3 - ... - 2^n + n2^(n+1)

= n2^(n+1) - 2[1 + 2 + ... + 2^(n-1)]

= n2^(n+1) - 2[2^n - 1]/(2-1)

= n2^(n+1) - 2^(n+1) + 2

= 2 + (n-1)2^(n+1).

B(n) = b(1)+b(2)+b(3) + ... + b(n-1) + b(n)

= 1*2*2^1 + 2*3*2^2 + 3*4*2^3 + ... + (n-1)n2^(n-1) + n(n+1)2^n

2B(n) = 1*2*2^2 + 2*3*2^3 + ... + (n-1)n2^n + n(n+1)2^(n+1),

B(n) = 2B(n) - B(n) = -2*1*2^1 - 2*2*2^2 - 2*3*2^3 - ... - 2*n*2^n + n(n+1)2^(n+1)

= n(n+1)2^(n+1) - 2[1*2^1 + 2*2^2 + 3*2^3 + ... + n*2^n]

= n(n+1)2^(n+1) - 2C(n)

s(n) = a(1)+a(2)+...+a(n) = b(1)+b(2)+...+b(n) - c(1) - c(2) - ... - c(n) = B(n) - C(n)

= n(n+1)2^(n+1) - 2C(n) - C(n)

= n(n+1)2^(n+1) - 3C(n)

= n(n+1)2^(n+1) - 3[2 + (n-1)2^(n+1)]

= n(n+1)2^(n+1) - 6 - 3(n-1)2^(n+1)

= [n^2 + n - 3n + 3]2^(n+1) - 6

= [n^2 - 2n+3]2^(n+1) - 6