One to a number series problem A number series problem

I don't know if the proportional series means, if so, the solution is as follows:

Because it is an equal proportional series, and the first term is a1=2 and the common ratio q=3, an=2*3 (n-1).

and bn=a3n-1, then bn=2*3 (3n-1)-1 (substituting n=3n into an can be done).

Solution: b(n+1) bn=a[3(n+1)-1] a(3n-1)=a(3n+2) a(3n-1)=3q=9

and b1=a(3 1-1)=a2=a1 q=6bn is the proportional series with the first term 6 and the common ratio 9.

bn=6×9（n-1）=54（n-1）(n＞1)

bn=a3n-1 This place is not clearly written, I see it as bn=(a3n) -1 solution.

bn+1=a3n=2*3 (3n-1)=(2 3)*27 n, so bn=(2 3)*27 n-1

The first 3 answers are all different, LZ still listen to me.

a1=2q=3

bn=a(3n-1)

So b(n+1)=a[3(n+1)-1]=a(3n+2)=27a(3n-1)=27bn

So bn is a proportional series of common ratio 27.

And because a1=2, a2=6

So when n=1, b1=a2=6

So bn=6*27 (n-1).

It's refreshing

Huh

**Unclear, can't see the question clearly.

Solution: Let the convex edge be an n-side.

The sum of the inner angles is (n-2)x180

The maximum angle is 160, and because the tolerance is 5, the minimum angle is 160-(n-1)x5=165-5n

The sum of the inner angles (n-2) x 180 = (160 + 165-5n) x n 2 so n = 9

The convex edge is a 9-sided shape.

Let the convex edge be an n-side.

n-2）x180

n-2)x180=(160+165-5n)x n2n=99-sided.

When n=1, a1=s1=3*1+5=8

When n 2 an=sn-s(n-1).

6n + 2 (tested a1=8 satisfied).

In summary, an=6n + 2

64b(n+1)-bn=0, i.e. b(n+1) bn= 1 64

So bn is a proportional series with the first term b1=8 and the common ratio q=1 64.

lz should know this: if an is a proportional series, then lg an is an equal difference series, so I'll prove it.

lg an -lg a(n-1)=lg an a(n-1)=lg q, the tolerance is lg q hehe, logcbn is an equal difference series).

then logcbn=logcb1 + (n-1)logc1 64

logc8 + (n-1) logc8 to the minus 2 power.

3-2n）logc8

Because there is a constant c, it is constant m for any natural number nan+logcbn (c is the base).

So an+logcbn=6n+2+(3-2n)logc8 should be eliminated in this equation n parameter, so it can be seen that logc8 should be equal to 3

Then c=2 so m=11

Oh, it's over, I'm just going to review the number series these days, I wish us progress together! =

Solution: The quadratic function f(x) x 2 ax a(a 0,x r) has and only one zero point.

i.e. there is only one solution to the equation x 2 ax a 0.

＝(a)^2－4a＝0

Solution: A 4 or A 0 (rounded).

f(x)＝x^2－4x＋4

The first n terms of the series and sn f(n).

sn＝n^2－4n＋4

When n 2, an sn s(n 1) 2n 5

When n 1, a1 s1 1 2 4 1 4 1

an＝1 (n＝1)

2n－5 (n≥2)

Use the dislocation subtraction method to find the sum of the first n terms of the series.

tn＝(1/3)＋[1)/(3^2)]＋1/(3^3)]＋3/(3^4)]＋2n－7)/(3^(n－1))]2n－5)/(3^n)]

1/3tn＝[1/(3^2)]＋1)/(3^3)]＋1/(3^4)]＋3/(3^5)]＋2n－7)/(3^n)]＋2n－5)/(3^(n＋1))]

, get: 2 3tn (1 3) [2 (3 2)] 2[(1 3 3) (1 3 4) ....1/3^n)]－2n－5)/( 3^(n＋1) )

tn＝1/3－[(n－1)/(3^n)]

Solution: an=lg 100sin n-1

4〗=lg〖100（√2/2)^(n-1)〗=lg100+(n-1)lg（√2/2)

2-(n-1)(lg2)/2

Because a(n+1)-an=2-(n+1-1)(lg2) 2-[2-(n-1)(lg2) 2]=-lg2) 2 is constant.

So {an} is a series of equal differences, with a tolerance of d=-(lg2) 2, and a1=2-(1-1)(lg2) 2=2

Let an=2-(n-1)(lg2) 2 0, get: n, that is, there are 15 positive terms, so the sum of the first 15 terms is the largest.

The maximum sum is s15=15a1+15(15-1)[-lg2) 2] 2=15*2-105lg2 2=30-105lg2 2

Let : when n is greater than or equal to 2, there is x that satisfies an+x=-3(a(n-1)-(5+x) 3)(1).

It can be obtained by deformation of an+x=-3a(n-1)+5+x, the purpose is to construct a proportional series, let bn=an+x, that is, the proportional series we want to construct is bn=-3*b(n-1)(2)}

Therefore, from the equations (1) and (2), x=-(5+x) 3 can be obtained, then: x=-5 4 is substituted x=-5 4 into bn=an+x to obtain bn=an-5 4, b1=3 4, so bn=(3 4) hole stool knows *(-3) (n-1) so an=(3 4)*(3) (n-1)+5 4 I don't know if there is a mistake in the calculation, but the method is absolutely correct. Hope it helps.

an=3an-1+5

an-1=3an-2=5

The upper formula minus the lower one.

an-an-1=-3 (an-1-an-2) treats an-an-1 as a number of waiters to roll rn

then rn=-3rn-1

rn=2(-3)^n-1

i.e. an-an-1 = 2(-3) n-1

an-1-an-2=2(-3)^n-2

a2-a1=2(-3)^0=2

Add it all up:

an=2((-3)^n-1+(-3)^n-2+..1)+2

Set the parameters to make up the proportional series.

an+t=-3(a(n-1)+t) =an=-3a(n-1)-4t =>4t=5 =>t=-5 Shanye4

an-5 4} is a series of first-class spring ratios, the first term 3 amuses the wild 4, q=-3an-5 4=3 4*(-3) (n-1)> an=3 4*(-3) (n-1)+5 4

This question is a(n+1)=2*(1+2+.n)+3*n+a1=n*(n+4)+1

an=(n-1)(n+3)+1

If you don't want to look at it that way, you have to take it apart and make up the constant sequence.

Since a(n+1) and an are preceded by a coefficient of 1, only the higher order can be taken.

For example, an+1 - an=3

an+1 -3n=an -3(n-1) So an+1 -3n is a constant sequence.

an+1 -an =2n

an+1 -n(n+1)=an - n(n-1) So an+1 -n(n+1) is a constant sequence.

Combined, it is.

an+1 -n(n+1)-3n is an+1 -n(n+4) is a constant sequence.

Because a1-0(0+4)=1

So an+1 -n(n+4)=1

an+1=n(n+4)+1

an=(n-1)(n+3)+1

an=an-1+2n+1

an-1=an-2+2n-1

a2=a1+5

That is, an+1 -n(n+4) is a constant sequence.

Since a(n+1) and an are preceding coefficients of 1, only the higher order an=(n-1)(n+3)+1 can be taken

an=an-1+2n+1

an-1=an-2+2n-1

a2=a1+5

If you don't want to look at it that way, you have to take it apart and make up the constant sequence.

Since a(n+1) and an are preceded by a coefficient of 1, only the higher order can be taken, such as an=(n-1)(n+3)+1