Question

数列{an}的前n项和为sn=n²+n 1.求数列{an}的通项公式

2.若bn=2^n·an，求数列{bn}的前n项和Tn

Answer 1

Sn = n^2+n
n=1, a1=2
for n>=2
an = Sn -S(n-1)
=2n-1 +1
=2n
an =2n

追问

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追答

Sn = n^2+n
n=1, a1=S1 = 1+1 =2
for n>=2
an = Sn -S(n-1)
=n^2 +n -[(n-1)^2 +(n-1)]
=2n-1 +1
=2n
an =2n