急!!很難的積分題(贈10點)

a_n = =∫(0,1)x^n(1-x)²dx

= ∫(0,1) x^n(1 - 2x + x^2) dx

= x^(n + 1)/(n + 1) - 2x^(n + 2)/(n + 2) + x^(n + 3)/(n + 3) | [0,1]

= 1/(n + 1) - 2/(n + 2) + 1/(n + 3)

Σ(k=1,n)ak

= [(1/2 - 1/3) - (1/3 - 1/4)] + [(1/3 - 1/4) - (1/4 - 1/5)] + ... + [[1/(n + 1) - 1/(n + 2)] - [1/(n + 2) - 1/(n + 3)]]

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

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

= (n^2 + 5n + 6 - 6n - 18 + 6n + 12)/6(n + 2)(n + 3)

= n(n + 5)/6(n + 2)(n + 3)

(2) lim(n→∞) n³ * an

= lim(n→∞) n³ * 1/(n + 1) - 2/(n + 2) + 1/(n + 3)

= lim(n→∞) n³ * [[1/(n + 1) - 1/(n + 2)] - [1/(n + 2) - 1/(n + 3)]]

= lim(n→∞) n³ * 1/(n + 1)(n + 2) - 1/(n + 2)(n + 3)

= lim(n→∞) n³ * 2/(n + 1)(n + 2)(n + 3)

= lim(n→∞) 2/(1 + 1/n)(1 + 2/n)(1 + 3/n)

= 2

設an=∫(0,1)x^n(1-x)²dx，(n=1,2,3,…)
(1)求Σ(k=1,n)ak
(2)求lim(n→∞)n³*an
[解]
(1) Σ{a_k: k=1,...,n}
　　 = Σ{∫_[0,1] x^k(1-x)^2 dx: k=1,...,n}
　　 = ∫_[0,1] (1-x)^2Σ{x^k: k=1,...,n} dx
　　 = ∫_[0,1] (1-x)^2 x(1-x^n)/(1-x) dx
　　 = ∫_[0,1] x(1-x)(1-x^n) dx
　　 = ∫_[0,1] (x-x^2-x^{n+1}+x^{n+2}) dx
　　 = 1/2-1/3-1/(n+2)+1/(n+3)
　　 = 1/6 - 1/[(n+2)(n+3)]
　　 = n(n+5)/[6(n+2)(n+3)](2) a_n = ∫_[0,1] (x^n-2x^{n+1}+x^{n+2}) dx
　　　 = 1/(n+1) - 2/(n+2) + 1/(n+3)
　　　 = 2/[(n+1)(n+2)(n+3)]
故
lim_{n→∞} n^3*a_n
　　 = lim_{n→∞} 2n^3/[(n+1)(n+2)(n+3)]
　　 = 2.Note: 若知 beta 函數, gamma 函數, 可用之.
　　 B(m,n) = ∫_[0,1] x^{m-1}(1-x)^{n-1} dx
　　　　　 = Γ(m)Γ(n)/Γ(m+n)