1. 若a1+2a2+3a3+…+nan=n(n+1)(n+2) 则an(通项公式)=
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1. 若a1+2a2+3a3+…+nan=n(n+1)(n+2) 则an(通项公式)=
2. 一个项数为偶数的等差数列,最后一项比第一项多10,且它的奇数项之和与偶数项之和分别是24和30,则这个数列共有几项
3. 求和:1/1*5+1/3*7+...+1/(2n-1)(2n+3)=
4. f(n)=1+1/2+1/3+...+1/3n-1(n属于N*)那么f(n+1)-f(n)=
2. 一个项数为偶数的等差数列,最后一项比第一项多10,且它的奇数项之和与偶数项之和分别是24和30,则这个数列共有几项
3. 求和:1/1*5+1/3*7+...+1/(2n-1)(2n+3)=
4. f(n)=1+1/2+1/3+...+1/3n-1(n属于N*)那么f(n+1)-f(n)=
![1. 若a1+2a2+3a3+…+nan=n(n+1)(n+2) 则an(通项公式)=](/uploads/image/z/19535806-46-6.jpg?t=1%EF%BC%8E+%E8%8B%A5a1%2B2a2%2B3a3%2B%E2%80%A6%2Bnan%3Dn%28n%2B1%29%28n%2B2%29+%E5%88%99an%28%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%29%3D)
1.
设bn=nan,其前n项和为Sn:
a1+2a2+3a3+…+nan=b1+b2+……+bn=Sn
即Sn=n(n+1)(n+2)
bn=Sn-S(n-1)=n(n+1)(n+2)-n(n-1)(n+1)=3n(n+1)
an=bn/n=3(n+1)
2.
设共有2n项,公差为d
a2n-a1=(2n-1)d=10
偶数项和-奇数项和=n(a2-a1)=nd=30-24=6
联立,得:n=3,2n=6
共有6项
3.
通项:1/(2n-1)(2n+3)=(1/4)·[1/(2n-1)-1/(2n+3)]
原式=(1/4)·[(1/1-1/5)+(1/3-1/7)+(1/5-1/9)+……+1/(2n-1)-1/(2n+3)]
=(1/4)·[4/3-1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/[(2n+1)](2n+3)]
4.
f(n)=1+1/2+1/3+……+1/(3n-1)
f(n+1)=1+1/2+1/3+……+1/(3n-1)+1/(3n)+1/(3n+1)+1/(3n+2)
f(n+1)-f(n)=1/(3n)+1/(3n+1)+1/(3n+2)
.
设bn=nan,其前n项和为Sn:
a1+2a2+3a3+…+nan=b1+b2+……+bn=Sn
即Sn=n(n+1)(n+2)
bn=Sn-S(n-1)=n(n+1)(n+2)-n(n-1)(n+1)=3n(n+1)
an=bn/n=3(n+1)
2.
设共有2n项,公差为d
a2n-a1=(2n-1)d=10
偶数项和-奇数项和=n(a2-a1)=nd=30-24=6
联立,得:n=3,2n=6
共有6项
3.
通项:1/(2n-1)(2n+3)=(1/4)·[1/(2n-1)-1/(2n+3)]
原式=(1/4)·[(1/1-1/5)+(1/3-1/7)+(1/5-1/9)+……+1/(2n-1)-1/(2n+3)]
=(1/4)·[4/3-1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/[(2n+1)](2n+3)]
4.
f(n)=1+1/2+1/3+……+1/(3n-1)
f(n+1)=1+1/2+1/3+……+1/(3n-1)+1/(3n)+1/(3n+1)+1/(3n+2)
f(n+1)-f(n)=1/(3n)+1/(3n+1)+1/(3n+2)
.
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