Question

已知等差数列an的公差为2,其前n项和Sn=pn^2+2n(n属于正整数)求p的值及an

Answer 1

an = a1+2(n-1)

Sn = n[a1+(n-1)]

Sn= pn^2+2n

coef of n^2

p=1

coef .of n

a1-1 = 2

a1 =3

an = 3+2(n-1) = 1+2n

bn = 2/[(2n-1)an]

Tn = b1+b2+..+bn

= summation ( 2/[(2n+1)(2n-1)] )

= summation ( 1/(2n-1) - 1/(2n+1) )

= ( 1- 1/3) +(1/3-1/5) +(1/5-1/7)+..+ (1/(2n-1) -1/(2n+1) )

= 1- 1/(2n+1) > 9/10

最小正整数n = 5