Question

(｀・ω・´)

想問這題
他的2（1+2...+n-1）+(nー1)1這裡是怎麼變成
an=n^2+4的～～

an-ant
EX. 已知數列<a>滿足 q =5,anti=an+(2n+1)中n≥1,求其一般項a=?
an-an+ = 2(n-1)+1
an-1-an-z = 2(1-2) +.)
a2-a₁ =
4
= -24-1
anti-an= (2+1)
0²₂-a1 = 2(1) + |
an-a₁ = 2[¹+2+₁ ++)] + (n-11-1
an = n²³4 for nzl