Question

求數(shù)列的前n項(xiàng)和：1+1，

1

a

+4，

1

a2

+7，…，

1

an-1

+3n-2，…．

Answer 1

設(shè)Sn=(1+1)+(

1

a

+4)+(

1

a2

+7)+…+(

1

an-1

+3n-2)
將其每一項(xiàng)拆開再重新組合得Sn=(1+

1

a

+

1

a2

+…+

1

an-1

)+(1+4+7+…+3n-2)
當(dāng)a=1時(shí)，Sn=n+

(3n-1)n

2

=

(3n+1)n

2

當(dāng)a≠1時(shí)，Sn=

1- 1 an

1- 1 a

+

(3n-1)n

2

=

a-a1-n

a-1

+

(3n-1)n

2