Analyze the following series of questions in detail, and the detailed questions are as follows

The law of the sequence is that each term is twice as large as the previous term, and it is an equal proportional sequence. The general formula is an=2 (n-1).

The sum of the first 100 terms can be summed by the formula of proportional series: s=a1*(1-q (n-1)) (1-q)=1*(1-2 (100-1)) (1-2)=2 99-1

It's a proportional series!

Take AS in Flash as an example:

var sum=0;

var j=1;

for(var i=1;i<100;i++)for(var n=0;nj=j*2;

sum =sum+j;

trace(sum);

I've tried it with as.,It's okay to loop to 10.,But to 100 the number is too big and it seems to overflow.,You can change the definition of sum.,Change it to double precision to see if it's okay!

The correspondence is as follows.

2 to the zero power, once square, 2 to the second power, to the third power, to the fourth power...

The sum of the first 100 terms is summed by a series of proportional numbers, which should be known.

The first item is 1, the common ratio is 2, a total of 100 items...

(n-1) power of an=2.

sn=(2 to the nth power -1).

So s100 = 2 to the power of 100 -1

an=2^(n-1)

Programming? What kind of program is it?

Let the general term of the proportional series be q, then a1+a2....a6 is the sum of the first six terms, so s6 = a1 (1-q 6) (1-q) = 1, (1).

1/a1+1/a2+..1 a6 is the sum of the first 6 terms of the proportional series with 1 a1 as the first term and 1 q as the common ratio, so 1 a1*(1-(1 q) 6) (1-1 q)=10,(2).

Divide (2) by (1) to get 1 (a 2q 5) = 10;i.e. a 2q 5 = 1 10, a1*a2*.a6=a1*a1q*..a1q^5=(a1)^6*q^15=（a^2q^5）^3=1/1000

The analysis and solution process is as follows: (k, n are integers) It is easy to see the law of the number series: The items with the value of k have k terms. 1 2 . n

1) Since p is the number of perennial rot and hunger, a1=1, when n=1, 2sn=2a1=2=p(2*1+1-1)=2p, so, p=1

sn=n(an+a1) pin2=n(an+1) 2

2sn=n(an+1)=2an +an-1, and, an>0, so, an=(n+1) 2

2).bn=an/2^n=(n+1)/2^(n+1)

tn=b1+b2+……bn

2/2^2+3/2^3+……n/2^n+(n+1)/2^(n+1)

2tn=2/2+3/2^2+4/2^3+……n+1)/2^n

tn=2tn-tn

2/2+1/2^2+1/2^3+……1/2^n-(n+1)/2^(n+1)

1/2+(1/2+1/2^2+……1 Hunger Return 2 n)-(n+1) 2 (n+1).

5/2-(n+5)/2^(n+1)

(1)(i)

an=a1+(n-1)d1

bn =b1+(n-1)d2

an+bn= (a1+b1)+(n-1)(d1+d2)=> is a series of equal differences, tolerance = d1+d2, first term = a1+b1(ii).

an = 1 proportional series.

bn = 2 n proportional series.

an+bn = 1+2^n

It's not a proportional series.

iii)an =a1+(n-1)d

bn = a(2n-1)

a1 +(2n-2)d

a1 + n-1)(2d)

is a series of equal differences, tolerance = d1 + d2 , first term = a1

Are you sure the answer is b? Not c??

Solution: 1-9 total of 9 numbers.

10-99 is a total of 90*2=180 numbers.

100-699 total 600 * 3 = 1800 numbers.

That is, a total of 9 + 180 + 1800 for a total of 1989 numbers.

i.e. the next number is the number sought.

So 700 is decomposed into 7,0,0

That is, the number on the 1990th digit is 7

If you don't understand, you can ask!

1990 is a prime number to find from prime numbers.

∵an=nb*a(n-1)/[a(n-1)+2n-2]

1/an=[(a(n-1)+2n-2]/nb*a(n-1)=1/nb+2(n-1)/nba(n-1)

nb/an=2(n-1)/a(n-1)+1

Let n an=cn rule.

bcn=2c(n-1)+1

cn=2/bc(n-1)+1/b=1/b[2c(n-1)+1]

cn+1/(2-b)=2/b[c(n-1)+1/(2-b)]

It is a proportional series with c1+1 (2-b)=1 a1+1 (2-b)=2 b(2-b) as the first term, and 2 b as the common ratio.

cn+1/(2-b）=2/b(2-b)*（2/b)^(n-1)=(2-b)*(2/b)^n

cn=)=2-b)*(2/b)^n-1/(2-b)=[2-b)^2*2^n-b^n]/(2-b)*b^n

i.e. n an=[(2-b) 2*2 n-b n] (2-b)*b n

an=n*(2-b)*b^n/[(2-b)^2*2^n-b^n]

The second question is proved by mathematical induction.

This mainly uses the construction method of the number series, the commutation method, the formula method, etc. I hope it will help you!

The biggest feature of this kind of problem is that the denominator on the right side of the equation is two parts and one part of the numerator. First of all, I want to take the reciprocal on both sides at the same time (note that the denominator cannot be 0, it must be checked). After collation, you will find that b(n an)=2(n-1 an-1)+1, let tn=n an, construct a new series, b*tn=2*(tn-1)+1, using the pending coefficient method.

You can construct a new proportional series. The problem is solved.

The first question is to take down, structure.

The second question is b=2 and bno=2 bno=2, and bno=2 can be done with mean inequality. Thank you.