The nth power of () is the 2nth power of a and the 3nth power of B

The nth power of () is the 2nth power of a and the 3nth power of B

The nth power of (2a & sup2; B & sup3;) = the 2nth power of a and the 3nth power of B

9 times the 3n-2 power of a times the 2n + 3 of B, which is a binomial of degree six
To be precise, if not, it can only be closed

Because it's six times
So the times of a plus the times of B = 6
That is: 3n-2 + 2n + 3 = 6
5N=5
N=1

M power of a = 2, n power of a = 3, find M-N power of a, 3N power of a, M + 2n power of A

a^(m-n)=a^m/a^n=2/3
a^3n=(a^n)^3=3^3=27
a^(m+2n)=a^m*(a^n)^2=2*3^2=18

In the sequence {53-3n}, what is the maximum number of terms n of the first n terms and Sn?

The general formula of sequence {53-3n} is as follows:
an=53－3n
Then:
a17>0、a18

The first n terms and Sn = - 3N ^ 2 + 17N (n belongs to N +) (1) the first n terms and the maximum (2) the general term formula an

1. Sn = - 3N ^ 2 + 17N (n belongs to N +)
Sn=-3(n-17/6)^2+289/12
When n = 3, there is a maximum value of Sn = 24
2.Sn=-3n^2+17n
S(n-1)=-3(n-1)^2+17(n-1)
Sn-S(n-1)=an=-6n+20

How to prove (1-1 / 3) (1-1 / 3 ^ 2) (1-1 / 3 ^ 3) (1-3^n)>1/2?

(1-1/3)(1-1/3^2)
=1 - 1/3 - (1-1/3) * 1/3^2
>1 - 1/3 - 1 * 1/3^2
=1 - 1/3 - 1/3^2
Similarly, if n times are processed, the result is as follows
(1-1/3)(1-1/3^2)(1-1/3^3)… (1-3^n)
>1 - 1/3 - 1/3^2 - 1/3^3 … - 1/3^n
=2 - ( 1 + 1/3 + 1/3^2 + 1/3^3 … + 1/3^n )
=2 - 1/(1 - 1/3) + 1/3^(n+1)
=2 - 3/2 + 1/3^(n+1)
=1/2 + 1/3^(n+1)
>1/2
I haven't read in vain for more than ten years, haha

It is proved that: 1 + 1 / √ 2 + 1 / √ 3 + 1 / √ 4 + +1√n＜2√n
It is suggested that when I ＞ 1, √ I + √ (i-1) ＜ 2 √ I, thus 1 / √ I ＜ 2 (√ I - √ (i-1))

One part of root K is less than two parts of the sum of root K and root k-1 = 2 (root k-root (k-1))
So the original formula is less than 1 + 2 (radical 2-radical 1) + 2 (radical 3-radical 2) + +2 (radical n-radical n-1) = 2 times the radical n-1, less than 2 times the radical n-1

Proof √ n

√n=1/[√n+√(n-1)]=√n-√(n-1)
1+1/√2+1/√3+...1/√n>=√n-√(n-1)+√(n-1)-√(n-2)+…… +√2-√1+1=√n

N ∈ R + prove 1 + 1 / 2 ^ 2 + 1 / 3 ^ 2 +... + 1 / N ^ 2 by scaling method

Certification:
n> 1:00
1/n²

Proof of 1 / 2-1 / (n + 1)

I guess you have the wrong title. I'll change it myself. Change 3 ^ 3 to 3 ^ 2
1/(2^2)+1/(3^2)+````+1/(n^2)
> 1/(2*3)+1/(3*4)+.+1/[n(n+1)]
=1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1/2-1/(n+1)
Right half
1/(2^2)+1/(3^2)+````+1/(n^2)
< 1/(1*2)+1/(2*3)+.+1/[(n-1)n]
=1-1/2+1/2-1/3.+1/(n-1)-1/n
=(n-1)/n