问题 解答题

设数列{an}中,an=1+2+3+…+n(n∈N*),将{an}中5的倍数的项依次记为b1,b2,b3,…,

(I)求b1,b2,b3,b4的值.

(II)用k表示b2k-1与b2k,并说明理由.

(III)求和:b1+b2+b3+…+b2n-1+b2n

答案

(I)∵an=1+2+3+…+n=

n(n+1)
2

由题意可得,b1=a4=10,b2=a5=15,b3=a9=45,b4=a10=55;

(II)∵an=

n(n+1)
2
=5m(m∈N+),

∴n=5k或n+1=5k(k∈N+),

即n=5k-1或n=5k

∵b2k-1<b2k

b2k-1=a5k-1=

5k(5k-1)
2
,b2k=a5k=
5k(5k+1)
2

(III)由(II)可得,b2n-1+b2n=

5n(5n-1)+5n(5n+1)
2
=25n2

∴b1+b2+…+b2n=(b1+b2)+(b3+b4)+…+(b2n-1+b2n

=25×12+25×22+…+25n2

=25(12+22+…+n2

b1+b2+…+b2n=

25
6
n(n+1)(2n+1).

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