考點(diǎn):數(shù)列的求和
專題:等差數(shù)列與等比數(shù)列
分析:當(dāng)n=2k-1(k∈N*)時(shí),a2k+1=a2k-1+1,數(shù)列{a2k-1}為等差數(shù)列,a2k-1=k;當(dāng)n=2k(k∈N*)時(shí),a2k+2=2a2k,數(shù)列{a2k}為等比數(shù)列,a2k=2k.分別利用等差數(shù)列與等比數(shù)列的前n和公式即可得出.
解答:
解:當(dāng)n=2k-1(k∈N
*)時(shí),a
2k+1=a
2k-1+1,數(shù)列{a
2k-1}為等差數(shù)列,a
2k-1=a
1+k-1=k;
當(dāng)n=2k(k∈N
*)時(shí),a
2k+2=2a
2k,數(shù)列{a
2k}為等比數(shù)列,
a2k=2k.
∴該數(shù)列的前16項(xiàng)和S
16=(a
1+a
3+…+a
15)+(a
2+a
4+…+a
16)
=(1+2+…+8)+(2+2
2+…+2
8)
=
+
=36+2
9-2
=546.
故答案為:546.
點(diǎn)評(píng):本題考查了等差數(shù)列與等比數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式、“分類討論方法”,考查了推理能力與計(jì)算能力,屬于中檔題.