[商管] [資結]-recurrence function
求解下列的recurrence function
題目：T(n)=T(n-2)+t(2)+3n , t(2)=1
Ans:
T(n)=T(n-2)+3*n+1
=[T(n-4)+3*(n-2)+1]+3*n+1
=[T(n-6)+3*(n-4)+1]+3*(n+(n-2))+2
=T(n-6)+3*(n+(n-2))+(n-4))+3
=...
=T(n-4)+3*(n+(n-2))+2
=T(2)+3*(n+(n-2))+(n-4)+....+4)+(n-2)
^^^^^^ ^^^^^^^^^^^劃線的地方我看不太懂
可以麻煩大家幫我解說一下嗎 感謝
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