High school math series problems, ask math geniuses to solve them

1.I open the square on both sides of the original formula and get it.

an+a(n+1)-1=2 [ana(n+1)], i.e. [ an- a(n-1)] =1

Monotonically increased by the question, a1>0, an- a(n-1)=1 an=n

i.e. an=n

For the natural number n 2 there are n >n -n

So 1 n < 1 [n(n-1)]=1 (n-1)-1 n1 2) +1 3) +1 4) +1 n) <1-1 2+1 2-1 3+1 3-1 4+...1/(n-1)-1/n=1-1/n<1

So 1 1 + (1 2) +1 3) +1 4) +1 n) < 1+1-1 2+1 2-1 3+1 3-1 4+...1/(n-1)-1/n=2-1/n<2

i.e. 1 a1 + 1 a2 + ...1/an<2

bn=(n+1)/(n+3)^2

b(n+1)-bn

n 2-3n+2) [(n+4) 2(n+3) 2]<0 indicates that bn is a decreasing series.

i.e. b1>b2>b3>b4>...bn

And because b1 = 1 8

b2<1/8

b3<1/8

So b1 + b2 + b3 + b4 + .bn<1/8+1/8+1/8+1/8+..1/8=n/8(n>=2)

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The first question can be solved with a2=4, a3=9, a4=16. Using mathematical induction, the general formula for this series is an=n 2

Next, scale-down. Obviously, n*n>n*(n-1), so 1+1 4+1 9+1 n 2<1+1 (1*2)+1 (2*3)+....1 n*(n-1), the split terms are eliminated, and the original formula is equal to 1+1-1 n<2, and the proof is completed.

The second question, since an has been found, then we can get bn=(n+1) (n+3) 2, and then we don't know ......

<> second question I haven't figured it out yet.

I'm not a math genius, so...

Proof: Vedic theorem has an+an-1=p, an*an-1=q

There is a formula for summing according to the series of equal differences.

s2n=[a1+a(2n+1)]*2n/2=n[a1+a(2n+1)]=n[an+a(n-1)]=pn

That is, the sum of this series is pn

To prove that the sum of this sequence is the root of (lgx) 2-*lgx+(lgn+lgp) 2=[lgx-(lgn+lgp)] 2=0.

That is, it proves that PN is the root of LGX=LGN+LGP.

It is only necessary to prove that pn satisfies the equation so that the equation holds.

So we bring in lgpn-lgn-lgp=(lgn+lgp)-lgn-lgp=0, which means that pn is the root of lgx=lgn+lgp.

So the sum of this sequence of s2n=pn is the root of (lgx) 2-*lgx+(lgn+lgp)2=0.

Solution: (1) In 2sn=a(n+1)-2 (n+1)+1, let n=1 obtain: 2s1=a2-2 2+1, let n=2 obtain: 2s2=a3-2 3+1, solution: a2=2a1+3, a3=6a1+13 and 2(a2+5)=a1+a3

The solution yields a1=1

2) A(n+2)=3a(n+1)+2 (n+1)+1 from 2sn=a(n+1)-2 (n+2)+1 a(n+2)+1 is obtained, and a1=1, a2=5 also satisfies a2=3a1+2 1, so a(n+1)=3an+2 n vs. n n* a(n+1)+2 (n+1)=3(an+2 n), a1=1,a1+2 1=3,an+2 n=3 n,an=3 n-2 n;

3）∵an=3^n-2^n=（3-2）（3^(n-1)+3^(n-2)×2+3(n-3)×2^2+…+2^(n-1)）≥3^(n-1)

1/an≤1/3^(n-1)

1/a1+2/a2+3/a3+..1/an≤1+1/3+1/3^2+..1/3^(n-1)= 1×[1-(1/3)^n]/(1-1/3)＜3/2

=1 2, the sequence satisfies f(1)=n 2*an(the square of n multiplied by an), then the general term of the series an = ? a1+a2+ +an=n^2*an a1+a2+ +a(n-1)=(n-1)^2*a(

Each item of the sequence is positive, an≠0, and both sides of the equation are divided by an 2[a(n+1) an] 2=2+[a(n+1) an][a(n+1) an] 2-[a(n+1) an]-2=0[a(n+1) an +1][a(n+1) an -2]=0a(n+1) an +1 constant "0", for the equation to hold, only a(n+1) an-2=0

a(n+1) an=2, which is a fixed value, and the number series is a proportional series with 2 as the common ratio.

a2+a4=2a3+4

2a1+8a1=8a1+4

a1=2an=2×2^(n-1)=2^n

The general formula for a series of numbers is an=2 n

Assuming that b1, bm, and bn are equally proportional, then.

bm^2=b1bn

m (2m+1)] 2=(1 3)[n (2n+1)]

4nm^2-(2n+3)m+n=0

To have a real root in the equation, discriminant equation 0

(2n+3)]^2-16n^2≥0

4n^2-4n-3≤0

2n+1)(2n-3)≤0

1 2 n 3 2, n is a positive integer, n can only be 1, and the equation becomes 4m 2-5m + 1 = 0

4m-1)(m-1)=0

m=1 4 (not an integer, rounded) or m=1 (m=n, rounded) in summary, there is no m,n that satisfies the topic.

Is it 1+ (n an) or (1+n) an ??

a(n) non-zero, a(1)] 2=[a(1)] 3,1=a(1)

a(n+1)]^3=[a(1)+a(2)+.a(n+1)]^2-[a(1)+a(2)+.a(n)]^2=a(n+1)[a(n+1)+2a(1)+2a(2)+.

2a(n)],a(n+1)]^2=a(n+1)+2a(1)+2a(2)+.2a(n),a(2)] 2=a(2)+2a(1)=a(2)+2, 0=[a(2)] 2-a(n)-2=[a(2)-2][a(2)+1], a(2)=2 or a(2)=-1

a(n+2)]^2=a(n+2)+2a(1)+2a(2)+.2a(n)+2a(n+1),a(n+2)] 2-[a(n+1)] 2=a(n+2)-a(n+1)+2a(n+1)=a(n+2)+a(n+1),0=[a(n+2)+a(n+1)][a(n+2)-a(n+1)-1],a(n+2)=-a(n+1) or a(n+2)=a(n+1)+1, Let's take a look at the various possible values for such a series:

a(1)=1

a(2)=2,-1

a(3)=3,-2,1

a(4)=4,-3,2,-1,a(5)=5,-4,3,-2,1,..a(2011)=2011,-2010,2009,..2,1,a(2012)=2012,-2011,2010,..2,-1

If n<=2011, a(n)=n, n>2011, a(n)=-a(n-1), then a(2012)=-a(2011)=-2011

When x=1, an is equal to bn

When x(1, 10, 1), a3b3Suppose that when n=k, an>bn, i.e., (1+lgx) k>1+klgx+k(k-1) 2(lgx) 2

Multiply both sides by (1+LGX) at the same time.

1+lgx)^(k+1)>1+(k+1)lgx+k(k+1)/2(lgx)^2+k(k-1)/2*(lgx)^3

Because lgx>0Therefore, (1+lgx) (k+1)>1+(k+1)lgx+k(k+1) 2(lgx) 2+k(k-1) 2*(lgx) 3>1+(k+1)lgx+k(k+1) 2(lgx) 2, i.e., a[k+1]>b[k+1], n=k+1;

So when x (1,+, an>bn.

From the title, it is clear that the common ratio of such a series of ratios will not be negative, nor will it be a number less than one. The former will not satisfy the requirement of the equal difference series, and the last term of the latter tends to zero, which is unreasonable.

Therefore, the common ratio is greater than one, so the equal difference series is incremental, that is, the tolerance is greater than 0.

and a5*a5=a3*an1, i.e. 36=a3*an1. an1>0 so a3 is also greater than 0. and an1>6>a3

If a3 is an integer, then the divisors of less than 6 in 36 are all divisors of 12, which can be proved.

1) The general formula for an: a(n)=a(n-1+1)=a(n-1)+a1=a(n-1)+2 So an-a(n-1)=2 shows that an is the first difference series with 2 The tolerance is 2 an=2n ;

2) an-a(n-1)=(-1) (n-1) bn (2 n+1) =2 is actually the last term of an, I can't see clearly, so just ball out of bn.

3) The third question, I am really inconvenient to play, there are several methods, you can give it a try: 1) c(n+1)-cn is greater than 0, which can mean that the horizontal is established, 2) division greater than 1 can mean that the horizontal is established.

3) I remember that I should have learned how to find the derivative of an exponent in high school, c1=3+6=9 Let's see cn as a function As long as the slope (and the derivative is a concept) is greater than or equal to zero and this function is a function that increases forever, then the above requirements can also be established You can just play the ball I think this is a little more reliable I really don't have a place for you to calculate it at work.

Let me give you some ideas.

1) Let p=n-1, q=1.It can be found that the number series an is an equal difference series. and an=2n

2) From an=2n, the tolerance is 2, according to the title: a1=b1 3,a2=b1 3-b2 5, (from this we can see that there will be one more formula after every n plus 1, that is, b1 3=2, -b2 5=2, and each term added later is equal to 2).

2=(-1) (n-1)*bn (2 n+1), so we can get bn=2*(2 n+1) (-1) (n-1) The third question is not clear.

I saw it on my phone, and I never understood that there was no master downstairs to answer?