The derivative of X * ln (x / 2)

The derivative of X * ln (x / 2)

=1*ln(x/2)+x*1/(x/2)*(x/2)'
=ln(x/2)+x*2/x*1/2
=ln(x/2)+1

Finding the derivative of y = ln (x ^ 2 + 1)

y＝ln(x²+1)
→y'＝(x²+1)'/(x²+1)
→y'＝2x/(x²+1).

-The derivative of (x + 2 / x) + ln (x)

-The derivative of (x + 2 / x) + ln (x)
=-1+2/x²+1/x

If BN = n power * an of 3, find the sum of the first n terms of BN
an=2n-1

bn=3^n*(2n-1)=2n*3^n-3^n
The sum of the first n items of 3 ^ n can be calculated
2n * 3 ^ n is a sequence of equal difference multiplied by equal ratio. The general method is to multiply by the ratio of equal ratio term, which is called 3, and then subtract by dislocation
First, let CN = 2n * 3 ^ n, 3CN = 2n * 3 ^ (n + 1), and then do it yourself

It is known that the general term sum of sequence an is n (n + 1), and the nth term BN of sequence BN is equal to the nth power of the 2nd of sequence an, that is, BN = the nth power of subscript 2 of A
1: Finding the general term an of tree sequence an
2: Finding the first n terms and Sn of sequence BN
3: For any positive integer n and K (KBN)

(1) Let the sum of the first n terms of an be TN = n (n + 1)
Then an = tn-tn-1 = n (n + 1) - (n-1) n = 2n
(2) Then BN = A2 ^ n = 4 ^ n
Sn=4（1-4^n)/1-4=4/3*（4^n-1)
(3) They are:
Bn-k + BN + k = 4 ^ (n-k) + 4 ^ (n + k) obviously, the two numbers are not equal, so the equal sign before the following radical is removed
>2 √ 4 ^ 2n (according to the nature of inequality: the mean of two numbers is greater than or equal to the geometric mean)
=2*4^n
=2bn,
The inequality that needs to be proved holds

It is known that n is a positive integer, and the two quadratic equations x ^ 2 + (2n + 1) + n ^ 2 = 0 of X are an, BN
It is known that n is a positive integer, and the two quadratic equations x ^ 2 + (2n + 1) + n ^ 2 = 0 of X are an and BN,
Find the value of the following equation:
1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+…… +1/(A20+1)(B20+1)

An+Bn=-(2n+1) An*Bn=n^2
Equation 1 / (an + 1) (BN + 1) = 1 / (an + BN + an * BN + 1) = 1 / N ^ 2-2n
=1 / N (n-2) = 1 / 2 [1 / (n-2) - 1 / N], so the formula is 1 + 1 / 2-1 / 19-1 / 20 = 531 / 380

It is known that n is a positive integer, and the two quadratic equations x ^ 2 + (2n + 1) + n ^ 2 = 0 of X are an and BN
It is known that n is a positive integer, and the two quadratic equations x ^ 2 + (2n + 1) + n ^ 2 = 0 of X are an and BN,
Find the value of the following equation:
1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+…… +1/(A20+1)(B20+1)

From the relationship between the root of quadratic equation and coefficient (that is, the Veda theorem), we can get the following conclusion
An+Bn=-(2n+1) ,An*Bn=n^2 ,
Therefore, 1 / [(an + 1) (BN + 1)] = 1 / (an * BN + an + BN + 1) = 1 / (n ^ 2-2n-1 + 1) = 1 / [n (n-2)] = 1 / 2 * [1 / (n-2) - 1 / n],
So, the original formula = 1 / (1 * 3) + 1 / (2 * 4) +. + 1 / (18 * 20)
=1/2*(1-1/3+1/2-1/4+1/3-1/5+.+1/18-1/20)
=1/2*(1+1/2-1/19-1/20)
=531/760 .

The first n terms of A1 = 1 an = 2an-1 + 2 (￣ nth power) an and xnbn = an + 1 / 2 (nth power + 1) - an / 2n power n ≥ 1
(1) The first n terms and xn of A1 = 1 an = 2an-1 + 2 (￣ nth power) an
(2) BN = an + 1 / 2 (nth power + 1) - an / 2n power n ≥ 1

BN = an + 1 / 2 is BN = (an + 1) / 2 or BN = an + (1 / 2)
These details are very important, and it is difficult to answer questions if they are not clear

The n-th power of {an} = 2N-1, {BN} = 2,
Let TN = A1 / B1 + A2 / B2 +. + an / BN (n belongs to N +). If the n-th power of TN + (2n + 3) / 2 - 1 / N < C (C belongs to Z) holds, the minimum value of C is obtained

First use the dislocation subtraction to get the sum, and then transform the constant value into the maximum value

LIM (n →∞) (a1 + A2 + a3 +) The proof of LIM (n →∞) an / N = 0

lim（n→∞）(a1+a2+a3+… an)/n=a
→lim（n→∞）(a1+a2+a3+… an)-n·a=0
→lim（n→∞）(a1-a)+(a2-a)+(a3-a)+… (an -a)=0
→lim（n→∞）an -a=0
→lim（n→∞）an/n=0