若数列{an}的通项公式an＝1 n*2+3n+2,则数列的前n项和Sn＝

an+1=an+2n推出an=an-1+2(n-1)...a2=a1+2累加得an=a1+2(2+3+4+...n-1)an=2+n(n-1)an=n^2-n+2(n>=1)

Sn=n^2*anSn-1=(n-1)^2*an-1Sn-Sn-1=an=n^2*an-(n-1)^2*an-1an(n^2-1)=an-1*(n-1)^2an-1/an=(n+1)/(n-1)a1/

解题思路：利用前n项和与第n项之间关系计算解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/incl

an+1=3an+2,a(n+1)+1=3an+3=3(an+1)数列{an+1}成等比数列q=3an+1=(a1+1)*3^(n-1)=3*3^(n-1)=3^nan=(3^n)-1

(1)∵an+1＝2an＋1∴an+1＋1＝2(an＋1)∴数列{an＋1}是等比数列∴an＋1＝(a1＋1)×2^(n－1)＝2^n∴an＝2^n－1(2)设am≤0am+1≥0∴2m－49≤02(

an=(1+2+...+n)/n=(1+n)*n/2n=（1+n）/2a(n+1)=(n+2)/2bn=1/an·a(n+1)=4/(n+1)(n+2)=2/(n+1)-2/(n+2)S(bn)=b1

取n和n-1,两等式互减再答：n等于1另算

a(1)=s(1)=(2/3)a(1)+1/3,a(1)=1.s(n)=(2/3)a(n)+1/3,s(n+1)=(2/3)a(n+1)+1/3,a(n+1)=s(n+1)-s(n)=(2/3)a(n

a(n+1)-an=a*(n+1)^2+n+1-an^2-n=2na+a+1当n≤4时,2na+a+1>0a>-1/(2n+1)≥-1/9当n≥8时,2na+a+1

此类题目采用累加法或迭代法∵an+1-an=3n（往下递推）∴an-an-1=3(n-1)an-1-an-2=3(n-2).a3-a2=3×2a2-a1=3×1以上格式左边＋左边=右边+右边左边相加的

a(n+1)=-an+3n-54a(n+1)+x(n+1)+y=-an+3n-54+x(n+1)+ya(n+1)+x(n+1)+y=-[an-(3+x)n+54-x-y]令x=-(3+x)y=54-x

an=10+lg(2^n)=10+nlg2a(n+1)=10+lg[2^(n+1)]=10+(n+1)lg2a(n+1)-an=lg2a1=10+lg2an=10+lg2+(n-1)lg2

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

n+Sn=2an,所以1+s1=2a1=2s1即s1=a1=1且n+1+S(n+1)=2a(n+1)相减得1+a(n+1)=2a(n+1)-2ana(n+1)=2an+1a(n+1)+1=2an+2=

a1=10^2=10^(2^1)a2=a1^2=10^4=10^(2^2)a3=a2^2=10^8=10^(2^3).an=10^(2^n)

a(n+1)-an=-2(n+1)^2+k(n+1)-(-2n^2+kn)=-4n-2+k由于数列{an}为递减数列,则对于任意的n∈N*总有a(n+1)-an≤0恒成立即：-4n-2+k≤0对于任意

（1）a(n+1)=3an/(2an+3)a1=1a2=3a1/(2a1+3)=3/5a3=3a2/(2a2+3)=3/7a4=3a3/(2a3+3)=3/9=1/3a5=3a4/(2a4+3)=3/

A_{n}+A_{n+1}-1=n*(A_{n+1}-A_{n-1})-------------------------1A_{n-1}+A_{n}-1=(n-1)*(A_{n}-A_{n-2})--

注:数学符号不好输入,你将就着看吧.等差数列的公差d=(An)-(An-1)这里只要能够证明这个d是个固定值不随N的变化而变化或常数就可以了而(An)-(An-1)=lg2^n-lg2^(n-1)=l

Sn是数列｛an｝的前n项和Sn=a1+a2+a3+……+anSn=(2^1+2×1-1)+(2^2+2×2-1)+(2^1+2×3-1)+……+(2^n+2×n-1)Sn=(2^1+2^2+2^3+