Ask a few series of questions, ask a series of questions

Question 1: The difference between 1, 2, 6, 15, and 31 neighbors is: 1, 4, 9, and 16, i.e.:

1 squared, 2 squared, 3 squared, 4 squared. And so on, ? The difference from 31 is 5 squared, i.e. 25, so?

for 56. Question 2: Should 25 be 24? Because: 6 2 2 2 2 2 24 3 3 3 3 60 4 4 4 4 4, 120 5 5 5 5, and so on. is 6 6 6 6, which is 210.

Question 3: I can't figure it out.

Let's get a general idea! 1. 1 + 1 * 1 = 2, 2 + 2 * 2 = 6, 6 + 3 * 3 = 15, 15 + 4 * 4 = 31 two, three,

It's not good!

There is a pattern to the third question.

That is: 1, 2, 3, 5, 7, 11, 13, 17 are all prime numbers, that is, there is no divisor except for the sum itself! So I suspect that number should be 5 11!

1,2,3,5,7,11,13,17 are all prime numbers, that is, there is no other divisor outside the one and itself, this is for sure, see if it helps the master!

The last number should be 7 37, right? These numbers should be written as:

The molecular law is obvious: 2 3 4 5 6 7

The denominator should be: 1 squared plus 1 2 squared plus 1 3 squared plus 1 4 squared plus 1 5 squared plus 1 6 squared plus 1

So I think the last number should be 7 37!

an=sn-s(n-1)

sn=n (sn-s(n-1))-n(n-1)n -1)sn=n s(n-1)+n(n-1)n+1) n*sn=n (n-1)*s(n-1)+1 bn=(n+1)/n*sn,b(n-1)=n/(n-1)*s(n-1)

bn=b(n-1)+1

b1=2*s1=2a1=1

bn=nn=(n+1)/n*sn

sn=n²/(n+1)

sn=12n-n^2=11n+n-n^2=n * 11 + n * n-1)/2 * 2)

This is in accordance with the formula for the sum of the n terms of the preceding sheds of the equal difference series.

Wherein: the first term a1 = 11

Tolerance d=-2

Let the last positive term be the x:

a1+(x-1)d=11+(x-1)*(2) 0x, take x=6

That is, the first 6 terms are positive, and the 7th term is negative

The sum of the first 6 terms: s6 = 12 * 6-6 2 = 36

The sum of the residual excitations of terms 7 to n = sn-s6 = 12 n-n 2-36 0 and the first n terms of the vertical sum socks tn=s6 + sn-s6 | 36 + 12n-n^2-36）】 n^2-12n+72

The first n terms of the known nucleus excitation column and sn=12n-n a1=s1=12-1=11

a2=s2-s1=12×（2-1）-2²+1²=9an=sn-sn-1=12n-n²-[12（n-1）-(n-1)²]12n-12（n-1）+[n-1)²-n²]=12+[(n-1-n)(n-1+n)]=12-2n+1=13-2n

an=13-2n, the number column is an equal difference series.

an=13-2n

When n 6, an =an, tn=sn=12n-n, when n lead takes 7, an =-an, tn=-sn+2s6=-12n+n modified+2 (12 6-6 )=n -12n+72

Question 1: The numerator and denominator on the right are divided by an, a(n+1)=1 and swiftly open(1 an+n 2).Then take the reciprocal at the left and right ends and then shift the phase, you can get 1 [a(n+1)-1 an]=1 n 2, and then you can add the left and right ends at the same time, and finally the mu is left with 1 a(n+1)-1 a1=1+2 2+...

n 2, the second question: just turn the virtual n into n-1, and then divide it!!

The first series feels like something is wrong, so let's check it again!

The second brings n=n+1 into a1*a2*a3....an*a(n+1)=(n+1) 2, and dividing the mu by the disturbance beats on both sides of the conditional equation yields a(n+1)=(1+1 n) 2, then an=(1+1 n-1) 2(n>1).

The last number in the first row is 1 squared, and the last number in the second row is the square ...... of 2

And so on. The last number of the forty-fifth line is 45 squared 2025, while the last number of the forty-fourth line is 44 squared 1936.

2011 is greater than 1936 and less than 2036, so it belongs to the forty-fifth line.

an-a(n-1)=3 (n-1), and the superposition method an=(an-a(n-1))+a(n-1)-a(n-2)) a2-a1)+a1 was used

3^(n-1)+3^(n-2)+.3 2 + 3 + 3 0c This is the sum of a proportional series.

3^(n-1))/2

Let's grab the only conditions.

Solution: an=3 (n-1)+a(n-1), i.e., an-a(n-1) =3 (n-1).

a2-a1=3^1

a3-a2=3^2

a（n-1）-a（n-2）=3^(n-2)an-a(n-1）=3^(n-1)

The above equations are summed up.

an-a1=3^1+3^2+……3^(n-1)an=(3^n-1)/2

Simple, an-a(n-1)=3 n-1... a3-a2=3^2，a2-a1=3。Adding these sequences can wait until the formula of the proof, and the specific process will be understood by yourself.

The sequence a[n]-3 n2 is a constant sequence made of.

a[n+1]-3 (n+1) 2=a[n]-3 n 2. If you are not sure, you can assume that the sequence bn=a[n]-3 n 2 then the above equation can be simplified to b(n+1)=bn, i.e. all terms bn are equal.

And because b1=a1-3 1 2=1-3 2=-1 2, bn is a constant sequence of all terms that are -1 2.

In short, the sequence of numbers that we get in that solution is a constant sequence is derived from a[n+1]-3 (n+1) 2=a[n]-3 n 2, and a1=1 is just used to find this constant, not because a1=1 is a constant series.

an+an+1-[a(n-1)+an]=d+d=2d(n is greater than or equal to 2), do you understand? Specializes in high school math.