The power of 1b is related to the magnitude of n

Think of n as the derivative of x.

1+b) n-1 to the power -1 0 monotonically increasing.

When n=1 is (1+b)-1 0

So (1+b) to the nth power is greater than n

From the limit, combined with the function image, the feasible domain interval is obtained.

1 (1+a n)+1 (1+b n)>=1*****> pass.

1+a n+1+b >=(1+a n)*(1+b n) and subtract the same term.

1>=a^nb^n

i.e. 1> = (ab) n

And ab=[(ab) 1 2] 2<=[(a+b) 2] 2=1, so ab<=1

Similarly. ab n<=1.

Let's start with an example:

6 7>7 6---6>(7 orange drain 6) 6 assigns a=n (n+1)>(n+1) n=b to verify c a=(n+1) (n+2)>(n+2) (n+1)=d, then c a (n+1)=[n+1) n] (n+1)=[1+1 n] (n+1).

d/b/(n+1)=[n+2)/(n+1)]^n+1)=[1+1/(n+1)]^n+1)

Apparently c a > d b

c=c a*a, is the large number multiplied by the round spike to the large number.

Therefore. So c>d

Inductively, it is true for all n>2: n to the n+1 power and to the nth power of (n+1).

When n 2 the former is big, when n gear sail guess 3 the latter is big.

Proof: Apparently both are positive.

n+1) n n (n+1)=(1+1 row type n) n nn n*, car age (1+1 n) n

If n is a positive integer.

Then 1,2, the original pure cavity = (tan 4-tanx) (1+tan 4tanx) to do the loss shirt dx

tan(π/4-x)dx

sin( 4-x) empty cos( 4-x) dx- sin( 4-x) cos( 4-x) d( 4-x) dcos( 4-x) cos( 4-x)ln|cos(π/4-x)| c,2,

When n=1, the power of n+1 of n is equal to the power of (n+1).

When n = 2, 2 3 = 8, 3 2 = 9, the power of n + 1 is less than the power of n (n + 1).

When n is greater than 3, the power of n+1 of n is greater than the power of (n+1).

Answer: bn n*(n+1) 2

Method: Elimination.

It is equivalent to -1 2+2 2-3 2+4 2++ factor the next two terms to get bn=(2-1)*(2+1)+(4-3)*(4+3)+(6-5)*(6+5)+.1+2+3+4+.

n=n*(n+1)/2

Thank you downstairs for the reminder to take the first old: when n=2n+1, n is a natural number, bn=-n 2+n*(n-1) 2; When n=2n, n is a natural number, bn=n*(n+1) 2

The inductive method is adopted, because the key knows n" 2

When n=3, the 4th power of 3 is greater than the 3rd power of 4, and when n=4, the 5th power of 4 is greater than the 4th power of 5.

Assuming that when n=k, the manuscript of k is greater than k+1 to the kth power, then when n=k+1, the same verification is obtained.