Question

13．已知數(shù)列滿足${S_n}=2{n^2}-n+1$，則通項(xiàng)公式a_n=$\left\{\begin{array}{l}{2，}&{n=1}\\{4n-3，}&{n≥2}\end{array}\right.$．

Answer 1

分析 根據(jù)數(shù)列通項(xiàng)公式與前n項(xiàng)和的關(guān)系進(jìn)行求解即可．

解答 解：當(dāng)n≥2時(shí)，a_n=S_n-S_n-1=2n²-n+1-[2（n+1）²-（n+1）+1]=4n-3，
當(dāng)n=1時(shí)，a₁=2-1+1=2，不滿足條a_n=4n-3，
則通項(xiàng)公式a_n=$\left\{\begin{array}{l}{2，}&{n=1}\\{4n-3，}&{n≥2}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{2，}&{n=1}\\{4n-3，}&{n≥2}\end{array}\right.$

點(diǎn)評(píng) 本題主要考查數(shù)列通項(xiàng)公式的求解，根據(jù)當(dāng)n≥2時(shí)，a_n=S_n-S_n-1是解決本題的關(guān)鍵．