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[(5/6)^n/(5/6)]*1/6)^n(5/6*1/6)^n*(6/5)

6/5)*(5/36)^n

5 36) n is an infinitely decreasing proportional series.

a1=5/36,q=5/36

So sum = a1 (1-q) = (5 36) (29 36) = 5 29 so the original formula = (6 5) * (5 29) = 6 29

Replace 5 6 with x, and rewrite the equation as:

1-x)*∑nx^

The latter is a maclaurin of 1 (1-x) 2, which converges to the original function at x=5 6, so the primitive.

1-x)*[1/(1-x)^2]

1/(1-x)

It can be obtained from a(n)+a(n-1)=4*3 n.

a(2)+a(1)=4*3^2

a(3)+a(2)=4*3^3

a(4)+a(3)=4*3^4

a(n)+a(n-1)=4*3^n

Subtract the above equation from bottom to top to get a(n)-a(1)=4*(3 n-3 (n-1)

Multiply both sides of the equation by -1 to get a(1)-a(n)=4(3 2+3 3+3 4+...).3^(n-1))-4*3^n

The brackets can be calculated using the formula of the sum of proportional sequences, and then because a(1) is already known, a(n) can be found.

In parentheses is a proportional sequence with the first term 9 and a common ratio of 3, which is summed by the equation of the equation Sum of the spheres.

3 n-9) 2 So a(1)-a(n)=2(3 n-9)-4*3 n, so a(n)=a(1)-2(3 n-9)-4*3 n

i.e. a(n)=1-2(3 n-9)-4*3 n= 19-2*3 (n+1).

The calculation is a bit urgent, maybe the calculation is wrong, but the method is absolutely correct, and it is best to calculate it slowly by yourself.

This is the famous Fibonacci sequence.

Derivation of the formula for the general term of the Fibona series].

Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21 ......

If f(n) is the nth term of the series (n n+). Then this sentence can be written in the form of the following:

f(1)=f(2)=1,f(n)=f(n-1)+f(n-2) (n≥3)

Apparently this is a linear recursive sequence.

Method 1 of deriving the general term formula: using the characteristic equation.

The characteristic equation for the linear recursive sequence is:

x^2=x+1

x1=(1+ 5) 2, x2=(1- 5) 2

Then f(n)=c1*x1 n + c2*x2 n

f(1)=f(2)=1

c1*x1 + c2*x2

c1*x1^2 + c2*x2^2

The solution yields c1 = 1 5 and c2 = -1 5

f(n)=(1 5)*[5 means root number 5].

Method 2 for deriving the general formula: ordinary method.

Let the constant r,s

such that f(n)-r*f(n-1)=s*[f(n-1)-r*f(n-2)].

then r+s=1, -rs=1

n 3, yes.

f(n)-r*f(n-1)=s*[f(n-1)-r*f(n-2)]

f(n-1)-r*f(n-2)=s*[f(n-2)-r*f(n-3)]

f(n-2)-r*f(n-3)=s*[f(n-3)-r*f(n-4)]

f(3)-r*f(2)=s*[f(2)-r*f(1)]

Multiply the above n-2 equations to get:

f(n)-r*f(n-1)=[s^(n-2)]*f(2)-r*f(1)]

s=1-r，f(1)=f(2)=1

The above formula can be simplified to:

f(n)=s^(n-1)+r*f(n-1)

Then: f(n)=s (n-1)+r*f(n-1).

s^(n-1) +r*s^(n-2) +r^2*f(n-2)

s^(n-1) +r*s^(n-2) +r^2*s^(n-3) +r^3*f(n-3)

= s^(n-1) +r*s^(n-2) +r^2*s^(n-3) +r^(n-2)*s + r^(n-1)*f(1)

s^(n-1) +r*s^(n-2) +r^2*s^(n-3) +r^(n-2)*s + r^(n-1)

This is the sum of the terms of the proportional series with s (n-1) as the first term, r (n-1) as the last term, and r s as the tolerance).

s^(n-1)-r^(n-1)*r/s]/(1-r/s)

s^n - r^n)/(s-r)

The solution of r+s=1, -rs=1 is s=(1+ 5) 2, r=(1- 5) 2

then f(n)=(1 5)*

Solution 1: (1) A1=2, An+1=AN

a2=a1=2，a3=a2=2，a4=a3=2，…，an=an-1=2，∴an=2

2）a1=-1，an+1=an+n^2

a2-a1=1^2，a3-a2=2^2，a4-a3=3^2，…, an-an-1=(n-1) 2, add up these equations to give an-a1=1 2+2 2+3 2+....+n-1）^2=（n-1）n（2n-1）/6

an=a1+n（n-1）（2n-1）/6=-1+n（n-1）（2n-1）/6

3）a1=0，an+1=3an+4

an+1 ten2=3an+6=3(anten2).

an+1 dec2) (an+102) = 3

a2 dec 2) (a1 decod 2) = 3, (a3 deci 2) (a2 deci 2) = 3 ,..., an-2) (an-1 decidal 2) = 3, multiply the above equation to get (an-102) (a1 decidal 2) = 3 (n-1).

An dec 2 = (a1 dec 2) 3 (n-1).

an=-2 dec. 2 3 (n-1).

Question 2: If the annual interest rate is 40%, the quarterly interest rate is 40% every 3 months 4=10%, 2 years for a total of 8 quarters.

Amount after 2 years = 1 10 8 (1+10%) 8 = 1 10 8

(a) a1=2,a2=(-1)*2,a3=(-1)^2 * 2，..an=(-1)^(n-1)*2

b) a1=-1,a2=-1+1^2,a3=-1+1^2+2^2，..an=-1+1^2+2^2+..n-1)^2=-1+1/6*(n-1)*n*(2*n-1)

c) a1=0 multiplied by 3 (n-1).

a2=3*a1+4 multiplies 3 (n-2).

a3=3*a2+4 multiplies by 3 (n-3).

a(n-1)=3a(n-2)+4 multiplies 3 1

a(n)=3a(n-1)+4 multiplies 3 0

After adding the left and right of n formulas, the left and right terms are approximated, and a(n)=4+3*4+3 2*4+.3^(n-2)*4 + 3^(n-1)*0

4*(3^(n-1)-1)/(3-1) = 2*(3^(n-1)-1)

The annual interest rate is 40%, and the interest rate is 10% every 3 months

Amount after 2 years = 1000000000 * 1+10%) 8 = 214358881

a(0)=cos(16) (here the half-width formula is applied) a(1)=cos(5 16) because (cos5 = 16(cos) 5-20(cos) 3+5cos).

a(2)=cos(5^2π/16)

And so on. a(n)=cos(5 n 16) set the period to x, then there are 5 x 16- 16=2 a after finishing.

5 x-1=32a (x and a are positive integers) to obtain x=8a(2010)=a(2)=cos(25 16).

Your topic hurts me...

[2+(2+2 The half-angle formula is used here.]

Calculate it yourself later.

Convert a(n+1) 2a(n)-2n+2n to a(n+1) 2a(n), when n=1,2,3...., you can find a(n) 2 (n+1).

Let the number be listed as an, then a1+a2+a3+a4=40;an-3+an-2+an-1+an=80；

Then there is 4*(a1+an)=120, and a1+an=30 is obtained

and (a1+an)*n 2=210

So n=14