Question

88

求解這一題😭😭😭😭

類題 1
類單 設數列a>滿足遞迴式{0.
fa=5
$005 200
an=
解
an=a+〔2(n-1)+1],n≥2
A₁ = 5
Lan-an-1 + [2 (n-1) + 1], n²z
試求一般項

Answer 1

✨ 最佳解答 ✨

qn

aₙ - aₙ₋₁ = [2(n-1)+1]
aₙ₋₁-aₙ₋₂= [2(n-2)+1]
⁝
+) a₂ - a₁ = [2(1)+1]
-------------------------------------
aₙ - a₁ = {2[(n-1)+(n-2)+⋯+1]+1×(n-1)}
aₙ=a₁+{2 × [n(n-1)]/2 + (n-1)}
aₙ=5+{n(n-1)+(n-1)}
aₙ=5+{(n+1)(n-1)}
aₙ=5+{n²-1}
aₙ=n²+4