Known sequence 1 2n 2n 2 1 Find the result of S1, S2, S3 2 conjecture Sn and prove it by mathemati

1) When n=1, 1 2n(2n+2)}=1, 8 n=2, 1 2n(2n+2)=1, 24 n=3, 1 2n(2n+2)=1 48

s1 = 1 8, s2 = 1 6, s3 = 3 162) from (1) s1 = 1 8, s2 = 1 6, s3 = 3 16 conjecture sn=n [4(n+1)].

Mathematical induction: (1) When n=1 is clearly true.

2) Suppose the equation holds when n=k, sk=k[4(k+1)], and when n=k+1, s(k+1)=sk+1 2(k+1)(2(k+1)+2).

k/[4(k+1)]+1/4(k^2+3k+2)=k/[4(k+1)]+1/4(k+1)(k+2)

k(k+2)+1)/4(k+1)(k+2)(k+1)^2/4(k+1)(k+2)

k+1)/[4(k+2)]

In summary, sn=n [4(n+1)] is true for any real number n.

So sn=n [4(n+1)].

That's how it should be!!

From the known: an=1 2[1 2n-1 (2n+2)], then s1=1 8, s2=1 6, s3=3 16, then sn=1 2[1 2-1 (2n+2)]. Then find (1):

When n=1, it holds. (2): Assuming that when n=k, is true, then sk=1 2[1 2-1 (2k+2), then when n=k+1, s(k+1)=sk+a(k+1)=1 2[1 2-1 (2n+4)], bringing a(k+1) into the above equation verification, is also true.

then the proof is completed.

The easiest way to solve this problem is to use the split term method, to split the general term into 1 4 (1 n-1 n+1) and then sum the front and back terms to eliminate. As for the method of number return, it is very easy to do it in strict accordance with the two steps of number return.

1/n(n+1)=1/n-1/(n+1)

So. s1=1/2

s2=1-1/2+1/2-1/3=1-1/3=2/3s3=1-1/2+1/2-1/3+1/3-1/4=1-1/4=3/4sn=1-1/(n+1)=n/(n+1)

Mathematical induction is available in textbooks.

Just take it one step at a time. Oh.

Let the general formula an=2 (n-1) of the proportional series, then the sum of the first n terms of the series is bn=2 n - 1

b1= 1b2 = 1+2

b3=1 + 2 + 4··

It is the number series that is sought.

then the sum of the first n terms is.

sn=(2 - 1)+(2²-1)+(2³ -1) +2^n - 1)

2+2²+2³+…2^n) -n=2(2^n - 1) -n

2^(n+1) -n - 2

This question is not difficult, mainly using the method of group summation and dislocation subtraction, and the solution is as follows:

sn=1+(1+2)+(1+2+2^2)+(1+2+2^2+2^3)+…1+2+…+2^(n-1)]=n*1+(n-1)*2^1+…+1*2^(n-1) (1)

2sn=n*2^1+(n-1)*2^2+…+1*2^n (2)

2)-(1)：sn=-n+2^1+2^2+…+2^(n-1)+2^n=-n+2^(n+1)-2

Each item is a proportional series.

Make each item a1, a2, a3, a4, a5 ,......a(n-1), the first term of an is set to a1=1

The second term is set to a2=1+2

The third term is set to a3=1+2+2 2

a（n-1）=（1-2^n-1）/（1-2）an=（1-2^n）/(1-2)

sn=1+(1-2^2)/(1-2)+(1-2^3)/(1-2)+…1-2^n）/(1-2)=（1-2）+（1-2^2）+…1-2^n）/(1-2)=/(1-2)=2^(n+1)-n-2

The formula for summing the proportional series is a=1+2+2 2+......2^(n-1)①∴2a=2+2^2+2^3+……2 n - get (2-1)a=(2+2 2+2 3+......2^n)-(1+2+2^2+……2^(n-1))=2^n-1

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

2sn＝2+（2+2∧2）+（2+2∧2+2∧3）+.2+2∧2+..2∧n）

sn＝1+（1+2）+（1+2+2∧2）+.1+2+2∧2+2∧（n-1））

sn-2sn＝..

I've been looking for this result.

This method is called dislocation elimination.

sn-s(n-1)=an=1+2+2^2+..2^(n-1)=2^n-1

s(n-1)-s(n-2)=a(n-1)=2^(n-1)-1..

s2-s1=2^2-1

s1=1 is added to the left and right sides.

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

So sn=2 (n+1)-2-n

2 (n+1)-2-n is preceded by 2 to the n+1 power, and you can't make a mistake.

Let sn=1 2+3 4+5 8+......2n-1) 2 n, then 2sn=1+3 boji2+5 4+7 8+(2n-1) 2 (n-1)sn=2sn-sn=1+(3 2-1 2)+(5 4-3 4)+(7 8-5 8)+....2n-1)/2^(n-1)-(2n-3)/2^(n-1)]-2n-1)/2^n=1+[1+1/2+1/4+……1 2 (n-2)]-2n-1) 2 n=1+2-1 2 (n-2)-(2n-1) 2 n=3-(2n+3) 2 n

n>=2) when the person respects n=1, s1=1 2, bring into the above formula, and it is true. So sn=3-(2n+3) 2 n

n>=1)

s1=1/(1×3)=1/3

s2=1/(1×3)+1/(3×5)=1/3+1/15=5/15+1/15=6/15=2/5

s3=1/(1×3)+1/(3×5)+1/(5×7)=1/3+1/15+1/35=2/5+1/35=14/35+1/35=15/35=3/7

Deformation: s1=1 3=1 (2 1 +1) s2=2 5=2 (2 2+1) s3=3 7=(2 3+1).

Conjecture: sn=n (2n+1).

Proof: When n=1, s1=1 (2+1)=1 3, which is equal to the calculated result, and the expression is true.

Suppose that the expression holds when n=k(k n, and k 1), i.e., sk=k (2k+1), then when n=k+1, s(k+1)=1 (1 3)+1 (3 5)+1/[(2k-1)(2k+1)]+1/[[2(k+1)-1][2(k+1)+1]]

sk+1/[(2k+1)(2k+3)]

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

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

2k^2+3k+1)/[(2k+1)(2k+3)]

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

k+1)/(2k+3)

k+1) [2(k+1)+1], the expression is also true.

In summary, sn=n (2n +1).

Conjecture: sn=n (n+1).

Proof: When n=1, s1=1 (1 2)=1 2=1 (1+1), the expression of the conjecture is true.

Suppose that the expression holds when n=k(k n, and k 1), i.e., sk=k (k+1), then when n=k+1, s(k+1)=1 (1 2)+1 (2 3)+1/[k(k+1)]+1/[(k+1)(k+2)]

sk+1/[(k+1)(k+2)]

k/(k+1)+1/[(k+1)(k+2)]=[k(k+2)+1]/[(k+1)(k+2)]=(k²+2k+1)/[(k+1)(k+2)]=(k+1)²/[(k+1)(k+2)]

k+1)/(k+2)

k+1)/[(k+1)+1]

The same is true for the expression.

In summary, the expression of sn is sn=n (n+1).

sn=n/(n+1) （1）

Step 1: s2 = 1 1 * 2 + 1 2 * 3 = 1-1 2 + 1 2-1 3 = 1-1 3 = 2 3

Satisfy equation (1).

Step 2: Suppose sn=n (n+1).

Then sn+1=sn+1 (n+1)*(n+2)=n (n+1)+1 (n+1)*(n+2)=1-1 (n+1)+1 (n+1)-1 (n+2)=1-1 (n+2).

n+1)/(n+2)

Satisfy equation (1).

Hence sn=n (n+1).