求下列各数列前n项和:(2-3x5) (4-3x5)
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an=1/[(2n+1)(2n+3)]=[(2n+3)-(2n+1)]/[2(2n+1)(2n+3)]=(2n+3)/[2(2n+1)(2n+3)]-(2n+1)/[2(2n+1)(2n+3)]=1/
An=Sn-S(n-1)=-(3/2)(n*n)+(205/2)n-{-(3/2)(n-1)^2+(205/2)(n-1)}=-3n+2对任意n,An
(1/3)n(3n+2)=(1/3)n(3n+3)-(1/3)n=n(n+1)-n/3=(1/3)[n(n+1)(n+2)-(n-1)n(n+1)]-(1/6)[n(n+1)-(n-1)n](1/3)
求a‹n›=(3n+1)(2^n/3)的前n项和S‹n›=(1/3)[(4×2)+(7×2²)+(10×2³)+(13×2̾
sn=3*3^1+5*3^2+.+(2n+1)*3^n①3sn=3*3^2+5*3^3+.+(2n-1)*3^n+(2n+1)*3^(n+1)②①-②-2Sn=Sn-3Sn=-2n*3^(n+1),因
用错位相减法:sn=1*3^1+3*3^2+5*3^3+.+(2n-1)*3^n3*sn=1*3^2+3*3^3+.+(2n-3)*3^n+(2n-1)*3^(n+1)-2sn=1*3^1+2*3^2
an=Sn-Sn-1=1/3n(n+1)(n+2)-1/3n(n+1)(n-1)=n(n+1)所以1/an=1/n(n+1)=1/n-1/n+1数列(1/an)的前n项和=1-1/2+1/2-1/3+
sn=1/2+2/4+3/8...n/2^nsn/2=1/4+2/8...+n/2^(n+1)两式相减,得sn/2=1/2+1/4+1/8...+1/2^n-n/2^(n+1)=1-1/2^n-n/2
S(n)=(n-1)×2^(n+1)+2解法一:S(n)=2^1+2×2^2+3×2^3+…+n×2^n=n×(2^1+2^2+2^3+…+2^n)-[2^1+2^2+2^3+…+2^(n-1)]-[
解题思路:数列前n项和解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.
Sn=3+2^nSn-1=3+2^n-1an=sn-sn-1=3+2^n-3-2^(n-1)=2^n-2^(n-1)=2*2^(n-1)-2^(n-1)=2^(n-1)
以第35项分界:前n项和Sn,n=35Tn(第n项的表达式)(-3/2)(2n-1)+205/2,n=35Tn=Sn-S(n-1)=(-3/2)(2n-1)+205/2解答到字数限制了
sn=3*1-1+2^1+3*2-1+2^2+.+3n-1+2^n=3*(1+2+.+n)-n+2^1+2^2+.+2^n=3n(n+1)/2-n+2*(1-2^n)/(1-2)=(3n^2+3n-2
n=1时,a1=s1=﹣3/2×1²+205/2×1=101n≥2时,an=Sn-Sn-1=(-3/2n²+205n/2)-[-3/2(n-1)²+205(n-1)/2]
【方法1:强行展开a(n)表达式】1+2+……+n=n(n+1)/21^2+2^2+……+n^2=n(n+1)(2n+1)/61^3+2^3+……+n^3=n^2(n+1)^2/41^4+2^4+……
这个用错位相消法(这类等差乘以等比的都是这样做)Sn=C1+C2+……+Cn(三分之一)XSn=(三分之一)XC1+……+nXCn(千万记得错一位)两式相减得(三分之二)XSn=…………(自己算吧记得
(一)当n为偶数时,Tn=-1^2+2^2-3^2+4^2.-(n-1)^2+n^2=3+7+11+.+2n-1=0.5*(3+2n-1)*(n/2)=0.5*n*(n+1)(二)当n为奇数时,Tn=
n为偶数时-1+4-7+10-.-(3n-5)+(3n-2)=(-1+4)+(-7+10)+...+[-(3n-5)+(3n-2)]=3×n/2=3n/2n为奇数时-1+4-7+10-13+...+(