It is known that the series an is positive, Sn 1 2 an 2 an

a1=s1=1/2(a1^2 +a1) 1/2 a1^2 +1/2 a1 -a1 =0 1/2 a1^2 -1/2 a1=0

1/2 a1(a1-1)=0 a1-1=0 a1=1

an=sn-sn-1 =1/2(an^2+an) -1/2(an-1^2+an-1)

1/2(an^2-an-1^2) -1/2 (an + an-1) =0

1 2(an + an-1)(an - an-1 - 1)=0 an -an-1 - 1 =0 an -an-1 = 1 The number column is an equal difference series with a tolerance of 1

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

1/(1+2+3) +1/[n(n+1)/2]

2[1/2 +1/2 -1/3 + 1/3 - 1/4 + 1/4 -1/5 +.1/(n-1) -1/n + 1/n - 1/(n+1) ]

2[1-1/(n+1)] =2n/(n+1)

a2=2 2 [2n (n+1)] =(n+1) n >1 (n belongs to n).

an=sn-sn-1=(an^2+an-an_1^2-an_1)/2

Eliminate an+an 1 and there is only an-an 1=1;

After that, let's do the math yourself, and you need more hands and brains to learn math well.

an=sn-s(n-1), so Yu Wang's sleepiness is 2sn=an+1 an=[sn-s(n-1)]+1 [sn-s(n-1)], so sn+s(n-1)=1 [sn-s(n-1)][sn+s(n-1)][sn-s(n-1)]=1, so (sn) 2-[s(n-1)] 2=1, so Lingshen[s(n-1)] 2-[s(n-2)] 2=1......(s2) 2-(s1) 2=1 plus (sn) 2-(s1) 2=n-1s1=a1s....

s1=a1=(a1 +1/a1 ) 2

Launch a1=1

s2 =a1+a2 =1+a2 = a2 +1/a2) /2

Roll out a2 = 2 -1

s2 =a1+a2+a3 =√2 +a3= (a3 +1/a3) /2

Rollout A3 = 3 - 2

Guess an = n - n-1).

The following is proved by mathematical induction.

When n=1 is true, it is clearly true.

Let n=k hold, i.e., ak = k - k-1).

then n=k+1.

s(k+1)=sk+a(k+1) =a(k+1) +1/a(k+1)] 2

Bring in SK with AK: (AK +1 AK) 2+A(K+1) =A(K+1) +1 A(K+1)] 2

Collated AK +1 AK+A(K+1)-1 A(K+1)=0

And ak +1 ak = k - k-1) +1 [ k - k-1)].

k -√k-1)+(k +√k-1)]

2 k So there is a(k+1)-1 a(k+1)+2 k = 0

a(k+1)]²2√k a(k+1)-1 =0

The solution is: a(k+1)= k+1) -k or a(k+1)=-k+1) -k (negative numbers rounded).

So n=k+1 when an = n - n-1) also holds.

So, the general formula for {an}: an = n - n-1).

Evidence: N=1, A1 leakage + A1=2S1=2A1A1 -A1=0

a1(a1-1)=0

a1 = 0 (discarded to touch) or a1 = 1

n 2, sn=(an +an) 2 s(n-1)=[a(n-1) +a(n-1)] 2

an=sn-s(n-1)=(an²+an)/2 -[a(n-1)²+a(n-1)]/2

an-a(n-1) -an-a(n-1)=0an+a(n-1)][an-a(n-1)]-an+a(n-1)]=0an+a(n-1)][an-a(n-1)-1]=0an+a(n-1)constant"0, so only an-a(n-1)-1=0an-a(n-1)=1, is a fixed value. A number series is a proportional series with 1 as the first term and 1 as the common ratio.

an=1+n-1=n

sn=1+2+..n=n(n+1)/2

sn-[an +a(n+1) talk] 4

n(n+1)/2-[n²+(n+1)²]4

1) 6sn = (an+1) (an+2)6s(n-1) = (a(n-1)+1)(a(n-1)+2) two-liter argument type wants to reduce 6an=an 2+3an+2-a(n-1) 2-3a(n-1)-2an 2-3an-a(n-1) 2-3a(n-1) = 0(an+a(n-1))(an-a(n-1))-3(an+a(n-1))=0(an+a(n-1))(an-a(n-1)-3)=0 because each item is a positive number, the noisy guess is an+a(n-1)>0, so an-a(n-1)-3=0, so an-a(n-1)=3, so it is an equal difference series, the tolerance of the stove is 3,6a1=(a1+1)(a1+2)=a1 2+3a1+2a1 2-3a1+2=0(a1-1)(a1-2)=0a1=1 or a1=2, so an=1+3(n-1)=3n-2 or an=2+3( n-1)=3n-1

n=1s1=a1=1 The equation is multiplied by an on both sides, and there is the equation an 2-2sn*an+1=0;The idea is to find the recursive relationship and discover the law. Solve the equation an=sn- sn 2-1 (sn>an, so discard the other) and an=sn-s(n-1); Sn 2-1=s(n-1) 2 is obtained, and the known shift term (an+1 an) 2 is brought into the equal difference series.

Guess an = n - n-1).

n=1.

Suppose n=k is true, there is ak=k- (k-1), sk=(ak+1 ak) 2= k

Then n=k+1, a(k+1)=s(k+1)-sk=[a(k+1)+1 a(k+1)] 2- k

The solution is that a(k+1)= (k+1)- k, n=k+1 is also true, so an= n- (n-1) is proved.

When n=1, 2s1=2a1=a1 +a1

a1²-a1=0 a1(a1-1)=0

a1 = 0 (each item is positive, rounded) or a1 = 1

At n 2, 2sn = an + an

2sn-1=a(n-1)²+a(n-1)

2sn-2sn-1=2an=an²+an-a(n-1)²-a(n-1)

an²-a(n-1)²-an-a(n-1)=0

an+a(n-1)][an-a(n-1)]-an+a(n-1)]=0

an+a(n-1)][an-a(n-1)-1]=0

All items of the sequence are positive, an+a(n-1)constant》0, for the equation to be true, only an-a(n-1)=1, is a fixed value.

A series is a series of equal differences with 1 as the first term and 1 as the tolerance.

an=1+n-1=n

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

bn=n×(1/2)^an=n/2^n

tn=b1+b2+b3+..bn=1/2^1+2/2^2+3/2^3+..n/2^n

tn/2=1/2^2+2/2^3+..n-1)/2^n+n/2^(n+1)

tn-tn/2=tn/2=1/2^1+1/2^2+1/2^3+..1/2^n -n/2^(n+1)

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

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

tn=2 -1/2^(n-1) -n/2^n

an>0

n=1. a1+1/a1=2s1=2a1

a1=1n>=2.

2sn=an+1/an=sn-s(n-1)+1/(sn-s(n-1))

2sn*(sn-s(n-1))=(sn-s(n-1))^2+1sn^2=s(n-1)^2+1

The series is a series of equal differences, the first term s1 2=1, and the tolerance is 1, so sn 2=n

sn=root number n

an=sn-s(n-1)=root-numbern-root(n-1), substituting n=1 also agrees.