The problem of the higher difference series to the process .

1) The three sides of a right triangle are a-d a, a+d .Then (a>0,d>0) finds the sinusoidal value sina of the angle (denoted a) corresponding to the side length a-d.

Right triangle, so (a+d) 2 = (a-d) 2 +a 2, arranged: a=4d, that is, the length of the three sides of the right triangle is: 3d, 4d 5d.is defined by trigonometric functions.

sina= 3d/(5d) = 3/5.

2) Find the general term formula for the series.

When n=3, calculate 1 a3+1=3 2 , when n=7, 1 a7 +1=2, let the tolerance be d, 2= 3 2 +4d to get d=1 8

So: 1 a8 +1 = (1 a7+1)+d= 17 8Note: Maybe I misunderstood the general formula of the series.

However, the formula for the general term of the equal difference series: am= an + (m-n)d can be used.

1.Let the three sides be a, a+d, a+2d, d>0, and a is the minimum angle.

So, there is, a square + (a + d) square = (a + 2d) square. (Pythagorean theorem) a+d)(a-3d)=0

Because d>0, a=3d

Then, the three sides are 3D, 4D, and 5D

That is, sina=3 5 [ Let d<0, the same is true for 3 5] 2, let d be the equal difference, (1 a3+1)+4d=(1 a7+1), and find d=1 8

1 a7+1)+d=(1 a8+1), a8=8 9

A typical (or its multiples) of a right triangle, the sine of the smallest inside angle is 3 5, I don't know if the difference series you are describing is or is.

If it's a series of equal differences, then there is.

1/[a(3)+1]=1/3

1/[a(7)+1]=1/2

1/[a(8)+1]=1/2 + 1/2-1/3)/(7-3)=13/24 => a(8)=11/13

If it's a series of equal differences, then there is.

1/a(3)+1=3/2

1/a(7)+1=2

1/a(8)+1=2 + 2-3/2)/(7-3)=17/8 => a(8)=8/9

If it's a series of equal differences, then there is.

1/a(2+1)=1/2

1/a(6+1)=1

1/a(7+1)=1+(1-1/2)/(6-2)=9/8 => a(8)=8/9

Question 1: Set a-d, a, a+d; Then there are:

a-d)*(a-d)+a*a=(a+d)*(a+d)*(a+d) solution a=4d;

then these three items are 3D, 4D, 5D;

How the sine of the smallest angle is...

1. Let the three sides of the right triangle be a-d, a, a+d, then (a-d) +a =(a+d), and the arranged a=4d, then the three sides are 3d, 4d, and 5d

The sine of the minimum angle is the opposite side ratio hypotenuse =

From 6a1+(n+12)d=24, a1=(24-(n+12)d) 6 (24-(n+12)d) 6+5d=4-(n+12)d, 6+5d=4+(5-(n+12) 6)d, is the fixed value.

d is a variable. 5-（n+12）/6=0

n=18

Compare 6A1+(N+12)D=24, S11=11A1+55D=11(A1+5D), if S11 is a fixed value, then A1+5D is a fixed value. 6 must be mentioned in 6a1+(n+12)d=24, so divide by 6 to get 6(a1+(n+12)d 6)=6*4.

It's been a long time since I've done high school math, and it's been a long time since I've reviewed it to figure it out, and I don't know if it's right or not. The black arrows indicate the subtracted items as well as the result. The format will just look at it.

The first question basic quantity method a4a5=(1+3d)(1+4d)=11, d is greater than 0, d=2 3,d=-5 4 rounded, an=1+2 3*(n-1)=2 3*n+1 3 second question, dislocation subtraction, general term = (2n+1)*3 (n-1) steps: expression, substitution, multiplication by 3 dislocation, up and minus, calculation and arrangement, the result general term = (kn+b)*q (n+t).

Conclusion【kn (q-1)+b (q-1)-k (q-1) 2】q (n+1+t)-[b (q-1)-k (q-1) 2*q (1+t), n*3 n

Remember to guarantee a perfect score.

The teacher did not plan the spatial layout, and after this method passed, not only could the girl find that the other party was more unreliable.

(1) Let the tolerance of the series be d,, the common ratio of the is q, and according to the meaning of the question, {(6+d)q=16(9+3d)q2=60 is solved to obtain {d=2q=2

an=3+(n−1)×2=2n+1,bn=2n−1(2)∵anbn=(2n+1)⋅2n−1

tn=3×20+5×2+7×22+…+2n+1)×2n−1①2tn=+3×2+5×22+7×23+…+2n−1)×2n−1+(2n+1)×2n②

tn=3+22+23+....+2n−(2n+1)×2n∴tn=(2n−1)×2n+1

Step D is a bit chaotic and roughly as follows: the first five terms and s5 = 34, the last five terms and the lifting mu s5'=146；sn=/2=234;The solution is n=13;The first seven terms are the middle terms of the positive and the forest, so 2a7=(s5+s5'5 So a7=18

n>=2

an=sn-s(n-1)=2n²-3n-2(n-1)²+3(n-1)=4n-5

a1=s1=-1

Also conforms to an=4n-5

then a1<0, and n>=2 has an>0,|an|=an, so |a1|=1

A2 to AN has n-1 terms.

and =(a2+an)(n-1) 2=(4n-2)(n-1) 2 plus |a1|=(4n -6n+4) 2=2n -3n+2 so n=1,tn=1

n>=1,tn=2n²-3n+2

The second one is obviously an=-4n+5

So a1>0

and n>=2,|an|=4n-5

Same as the first one.

So n=1, tn=1

n>=1,tn=2n²-3n+2

Substituting 1 into 3 yields: n=15 three, a(n),sn,(a(n)) 2 into a series of equal differences, so 2sn=an=an 2 sn=an(an 1) 2 (1 formula) s100=(a100 a1)100 2 (2

a1=s1 d=sn-s(n-1) is ok.

a1=2*1^2-3*1=-1

d=(2*n^2-3n)-(2(n-1)^2-3(n-1))=4n-5

I can't find out how much d is, so you can find s2=a1+a2 again, and find a2 directly and find d faster, d=a2-a1. Because you know that it is an equal difference series, you can directly subtract it.

1. The sum of the first 101 items is 1111

a1+a2+a3+a4+..a101=1111 is.

a1+d*0

a1+d*1

a1+d*2

a1+d*3

a1+d*100

101a1+d*(100 2*101)=1111 Substituting d=1 5 to get a1=1

Then a1+a6+a11+....+a96=

a1+5d*0

a1+5d*1

a1+5d*2

a1+5d*3

a1+5d*19

20a1+5d*(19*20/2)

20a1+190

Second, s9=18 is (a1+a9)*9 2=18

So a1+a9=4

So a1+a1+8d=4

So a1+4d=2 (formula 1).

sn=240 is (a1+an)*n 2=240 (2).

Because a(n-4)=30(n>9), an=a(n-4)+4d=30+4d, substituting this into 2 formulas, gets:

a1+30+4d)*n 2=240 (3 formulas).

Substituting 1 into 3 yields: n=15

Three, a(n),sn,(a(n)) 2 into a series of equal differences, so 2sn=an=an 2

sn=an(an+1) 2 (1 formula).

s100 = (a100 + a1)100 2 (2 formula).

Substituting n=100 into 1 gives :

s100 = (a100 + a1) a100 2 and s100 = (a100 + a1) 100 2

So a100=100, and a1=s1, substituting n=1 into 1, i.e., a1=a1(a1+1) 2 obtains.

a1 = 1, so s100 = 5050

Four. a1+a7=2

a1+a15=10

Subtract to get 8d=8, d=1, so a1=-2

So sn=a1n+n(n-1)d 2=n(n-1) 2-2n

Number series (sn) n=n 2-5 2

The first term of this series b1=-2 so tn=(-2+n 2-5 2)n 2=n(n-9) 4