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设F(n)=2+2^4+2^7+2^10+...+2^3n+1(n∈N+),则F(n)=

f(n)=2+2^4+2^7+2^10+...+2^3n+1 =2(1-2^n)/(1-8)=-2/7(1-2^n)=1/7*[2^(n+1)-2] f(n)=1/7*[2^(n+1)-2]

由题意知,观察指数1,4,7,…,3n+10该数列的通项公式为3n-2,而3n+10为数列的第n+4项∴f(n)是首项为2,公比为8的等比数列的前n+4项和,所以f(n)= 2(1- 8 n+4 ) 1-8 = 2 7 ( 8 n+4 -1) .故答案为: 2 7 ( 8 n+4 -1)

f(n+1)-f(n)=1/(3n+1) + 1/(3n+2)+ 1/(3n+3)- 1/(n+1)=1/(3n+1) + 1/(3n+2)- 2/(3n+3)>0

因为数列各项的指数是:1,4,7,10…是以1为首项,3为公差的等差数列,所以其通项为:1+3(x-1)令3n+10=1+3(x-1)?x=n+4.即求首项为2,公比为2 3 的等比数列的前n+4的和.∴S n =2+2 4 +2 7 +2 10 +…+2 3n+10 = 2(1- 2 3(n+4) ) 1- 2 3 = 2 7...

由题得:问题是求首项为2,公比为2 3 的等比数列的前n+4项的和.∴S n = 2[1- (2 3 ) n+4 ] 1- 2 3 = 2( 8 n+4 -1) 7 .故答案为: 2( 8 n+4 -1) 7 .

设f(n)=an+b 则有: f(n+1)=a(n+1)+b=an+a+b f(f(n)=f(an+b)=a(an+b)+b=a^2 *n+ab+b 所以,有: f(n)+f(n+1)+f(f(n))=an+b+an+a+b+a^2 *n+ab+b=(a^2+2a)n+a+ab+3b=3n+1 所以有: a^2+2a=3 (1) a+ab+3b=1 (2) 解(1)得: a=1或a=-3 如果a=1则...

1/(n+1)+1/(n+2)+1/(n+3)+······+1/(3n) <1/(n+1)+1/(n+1)+1/(n+1)+······+1/(n+1) =[3n-(n+1)]/(n+1) =(2n-1)/(n+1) <(2n+2)/(n+1) =2

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