Question

【題目】已知數(shù)列{an}為等差數(shù)列,a1=1,前n項(xiàng)和為Sn,數(shù)列{bn}為等比數(shù)列,b1>1,公比為2,且b2S3=54,b3+S2=16.

(Ⅰ)求數(shù)列{an}與{bn}的通項(xiàng)公式;

(Ⅱ)設(shè)數(shù)列{cn}滿足cn=an+bn,求數(shù)列{cn}的前n項(xiàng)和Tn.

Answer 1

【答案】(Ⅰ) an=2n﹣1,; (Ⅱ)Tn.n2+3(2n﹣1).

【解析】

(Ⅰ)利用等差數(shù)列的前項(xiàng)和公式和等比數(shù)列的通項(xiàng)公式列式解方程組解得和，進(jìn)一步可得和；

(Ⅱ)利用等差數(shù)列與等比數(shù)列的前項(xiàng)和公式分組求和可得.

(Ⅰ)∵數(shù)列{an}為等差數(shù)列,a1=1,前n項(xiàng)和為Sn,數(shù)列{bn}為等比數(shù)列,

b1>1,公比為2,且b2S3=54,b3+S2=16.

∴,

解得b1=3,d=2,

∴an=2n﹣1,;

(Ⅱ)∵cn=an+bn=(2n﹣1)+32n﹣1.

∴Tn=c1+c2+…+cn=[1+3+…+(2n﹣1)]+3(1+2+22+…+2n﹣1)

n2+3(2n﹣1).