A high school math problem with a series of equal differences

Solution: The sum of the first n terms of the sequence is sn=2n2

Rollout: an=sn-sn 1=2n 2-2(n-1) 2=4n-2 then a1=2 a2=6

Because a1=b1 so b1=2

Bring in b2(a2-a1)=b1 b2(6-2)=2 to find b2=1 2

Because is a proportional series.

Since b2 b1=1 4

So bn=(1 4) (n-1)*b1=2*(1 4) (n-1).

When n is greater than or equal to 2.

sn=2n²

sn-1=2（n-1）²

Subtract the two formulas. an=4n-2

n is equal to 1.

s1=a1=2

Satisfy the above formula. So an=4n-2

b1=a1=2

b2(a2-a1)=b1

So b2=(b1) 4

q=1 4bn=2 (1 4) to the n-1 power.

From sn=2n 2 to get s(n-1), to subtract to get an to get a1 b1 and then b2(a2-a1) = b1 to get b2

bn} is an equal proportional series, and if you know the common ratio, you can do it.

a1=s1=2=b1

a2=s2-s1=6

a2-a1=4

Because is a proportional series.

So q=b2 b1=1 (a2-a1)=1 4, so bn=(1 4) n*8

an=sn-s(n-1)=2=4n-2

Solution: a1+a2+......AM = A1 + A2 + AN can be true.

Since a9=0

s9=s8+a9=s8+0=s8

Let m=9 n=8 or m=8 n=9, and the above equation holds.

Hello, look**, ask for tolerance first. Then calculate that the value after that term is less than . Then count the sum.

a1=24, a5=12, then an=a1+(n-1)d:

a5=a1+4d

> 12=24+4d

> d=-3

That is, down 3 per day

Household operating income (1800 + 1000 * 3) = 4800, other income (1350 + 160 * 3) = 1830, and the per capita income in 2012 was 6630 yuan.

Solution: There is a question to obtain: family income a1=1800 d1=10002012 farmer's household operating income in the area a3=a1+2d1=1800+2000=3800

Other income b1=1350 d2=other income of farmers in the area in 1602012 B3=b1+2d2=1350+320=1670

In 2012, the per capita annual income of farmers in this region was 3800+1670=5470

Annual household income an=1800+(n-2009)*1000

Other income bn = 1350 + (n-2009) * 160

The per capita income of 2000 is s=1800+(2012-2009)*1000+1350+(2012-2009)*160

a3 + a8 = 10 According to the basic inequality, the product is up to 25. That is, the number column is a constant series.

From an bn=(7n+1) 4n+27 a21 b21=148 111

a21=(a1+a21)*21 2=a11*21 so a11=a21 21

In the same way, b21=(b1+b21)*21 2=b11*21, so b11=b21 21

So a11 b11 = 148 111

Is 4n+27 the denominator? It's still separate.

For example, the difference is listed as: a3 = a1 + 2a, a9 = a1 + 8a, a15 = a1 + 14a, a17 = a1 + 16a

Launched: a3 + a9 + a15 + a17 = 4a1 + 40a = 8 gets: a1 + 10a = a11 = 2

So a11=2

Let the difference be d then: (* sign means multiplier sign).

a3=a1+2*d

a9=a1+8*d

a15=a1+14*d

a17=a1+16*d

a1+2*d)+(a1+8*d)+(a1+14*d)+(a1+16*d)=8

4*a1+40*d=8

a1+10*d=2

Then a11 = a1 + 10 * d = 2