求和:Sn=1+4/5+7/5^2+.+3n-2/5^(n-1)
求和:Sn=1+4/5+7/5^2+......+3n-2/5^(n-1)
Sn=1+4/5+7/5^2+......+3n-2/5^(n-1)
1/5*Sn=1/5+4/5^2+7/5^3+......+(3n-2)/5^n
两式相减，
得4/5Sn=1+3/5+3/5^2+......+3/5^(n-1)-(3n-2)/5^n
得4/5Sn=1+3/4(1-1/5^(n-1))-3n-2)/5^n
Sn=5/4+3/5(1-1/5^(n-1))-(3n-2)4/25^n
……
求和；1+4/5+7/5……2+……+3n-2/5^n-1
Sn=1+4/5+7/5^2+…+（3n-2）/5^n-1
那么5Sn=5+4+7/5+10/5^2+…+3n-2/5^n-2
两式错位相减得4Sn=8+3*(1/5+1/5^2+……+1/5^n-2)-(3n-2)/5^n-1
括号内是等比数列
4Sn=8+3*1/5*(1-1/5^n-2)/(1-1/5)-(3n-2)/5^n-1
4Sn=8+[3*5^(n-1)-12n-7]/[4*5^(n-1)]
Sn=2+[3*5^(n-1)-12n-7]/[16*5^(n-1)]
求和1-3/2+5/4-7/8+...+[(-1)^(n-1)]*(2n-1)/2^(n-1)
Sn= 1-3/2+5/4-7/8+...+[(-1)^(n-1)]*(2n-1)/2^(n-1)
2Sn=2-3+5/2-7/4+......[(-1)^(n-1)](2n-1)/2^(n-2)
两式相加：
3Sn=2/2-2/2^2+......+[(-1)^(n-1)]*2/2^(n-2)+[(-1)^(n-1)]*(2n-1)/2^(n-1)
3Sn=[1-(-1/2)^(n-2)]/(3/2)+[(-1)^(n-1)]*(2n-1)/2^(n-1)
Sn=[2-(-1)^(n-2)/2^(n-3)]/9+[(-1)^(n-1)]*(2n-1)/3*2^(n-1)
2/3！+3/4！+4/5！+...+（n-1）/n！，求和。
（n-1）/n！=1/（n-1)!-1/n!
2/3！+3/4！+4/5！+...+（n-1）/n！
=[1/2!-1/3!+1/3!-1/4!+.....+1/（n-1)!-1/n!]
=1/2!-1/n!
=1/2-1/n!
1/(1*3)+1/(2*4)+1/(3*5)+…+1/(n-1)(n+1)求和
=1/2*(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+....
+1/(n-2)-1/n+1/(n-1)-1/(n+1))
=1/2*(1+1/2-1/n-1/(n+1))
=3/4-1/2n(n+1)
=（3n^2+3n-2）/4n(n+1)
求和Sn=1/(1*3)+1/(2*4)+1/(3*5)+...+1/(2n-1)(2n+1)
求和Sn=1/(1*3)+1/(2*4)+1/(3*5)+...+1/(2n-1)(2n+1)
分析：∵1/(1*3)=½（1-⅓）1/(2*4)=½（½-¼）以此类推
∴Sn=½×[1-⅓+½-¼+⅓-1/5+...+1/(2n-1)-1/(2n+1)]
=½×[1+½-1/2n-1/(2n+1)]
=3/4-[(4n+1)/4n(2n+1) ]
求和：Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
=1+1/1*3+1+1/3*5+...+1+1/(2n-1)(2n+1)
=n+(1/1*3+1/3*5+...+1/(2n-1)(2n+1))
=n+[(1-1/3)/2+(1/3-1/5)/2+...+(1/(2n-1)-1/(2n+1)/2]
=n+(1-1/(2n+1))/2
=n+(2n/(2n+1)/2
=n+n/(2n+1)
求和：Sn=1*1/2+3*1/4+5*1/8...2n-1/2^n
错位相乘法，懂么这是一个比较基础的题目啊
步骤 1 Sn=a
2 1/2Sn=b
3 sn-1/2sn=1/2sn=a-b
具体的自己去算，方法已经给出。
求和Sn=2²/1·3+4²/3·5+...+(2n)²/(2n-1)(2n+1)
解：
令an=(2n)²/(2n-1)(2n+1)
=1/[1-(1/2n)][1+(1/2n)]
=(1/2)*{[1-(1/2n)]+[1+(1/2n)]}/[1-(1/2n)][1+(1/2n)]
=(1/2)*{1/[1+(1/2n)] + 1/[1-(1/2n)]}
=(1/2)*[2n/(2n+1) + 2n/(2n-1)]
=(1/2)*{1-[1/(2n+1)]+1+[1/(2n-1)]}
=1 + (1/2)*[1/(2n-1) - 1/(2n+1)]
Sn
=a1+a2+...+an
=1*n + (1/2)*[(1/1-1/3)+(1/3 - 1/5)+(1/5 - 1/7)+......+1/(2n-1) - 1/(2n+1)]
=n + (1/2)*[1 - 1/(2n+1) ]
=n{1+[1/(2n+1)]}
