作业帮已知等差数列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
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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
若bn=2/(2n-1)an,记数列bn的前n项和为Tn,求使Tn>9/10成立的最小正整数n的值
若bn=2/(2n-1)an,记数列bn的前n项和为Tn,求使Tn>9/10成立的最小正整数n的值
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
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
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