Binomial theorem must see!

This is easy.

Let f(x) = (1+1/x)^x (I am assuming a different function whose domain is x=R)

lnf(x) = x + ln(1+1/x)
=> f'(x) = f(x)[1 - 1/(x+1)(x)]
= f(x)(x²+x-1)/x(x+1)

f'(x) > 0 for x≥2

So the function is increasing in the interval [2,∞)

So max value will be when n-->∞ = e (from limits chptr)

min value will be when n=2 = 2.25

So for all values of x (inclusive of integers) 2< f(x)<3

from bino theorem
(1+1/n)ⁿ=1+ n.1n + n(n-1)n²2! + n(n-1)(n-2)n³3! + ............. n(n-1)(n-2)..........[n-(n-1)]nⁿn!

=1+1+12!(1-1n) + 13!(1-1n)(1-2n) + 14!(1-1n)(1-2n)(1-3n)...............1n!(1-1n)(1-2n)(1-3n)......... (1-n-1n)
<1+1+12!+ 13!...............1n!
<1+1+12+12²+ 12³...........................12^n-1
=3-12^n-1

check here n ≥1 for this 2<f(n)<3

Let a = ⁿ C _r-1 , b = ⁿ C _r , c = ⁿ C _r+1
and d = ⁿ C _r+2

Therefore, a+b = ⁿ⁺¹ C _r , b+c = ⁿ⁺¹ C _r+1 , c+d = ⁿ⁺¹ C _r+2

a+ba = ⁿ⁺¹ C _rⁿ C _r-1 = n+1r ........................(i)

and b+cb = ⁿ⁺¹ C _r+1 ⁿ C _r = n+1r+1..............................(ii)

and c+dc = n+1r+2 ...............(iii)

Frm (i), (ii), and (iii); we get, aa+b , bb+c, cc+d are in A.P

Now Apply, A.M≥ G.M, we get
bb+c ≥ √ac(a+b) (c+d)

Therefore, { (bb+c) ² - ac(a+b) (c+d) } ≥ 0

Therefore, thus proved.

in post#2
f(x) = (1+1/x)^x

lnf(x) = x + ln(1+1/x)

How ??

shouldnt it be lnf(x) = x.ln(1+1/x)

Ans) f(x,y) = x²+3y²+2xy-6x-2y

f _x = 2x +2y - 6

f _y = 6y + 2x - 2

Now Put f _x = 0 ........ that implies x+y = 3 .................(i)

Now Put f _y = 0 ......... that implies x+3y = 1 ....................(ii)

On solving (i) and (ii), we get
x = 4 and y=-1 (Stationary points)

Now, r = f_xx = 2 ;
S= f _xy = 2 ;
and similarily t = f_yy = 6

Since rt - s ² = 12 - 4 = 8 > 0 and r > 0 , then the function f(x,y) has a minimum value at point (4, -1)
Therefore, f _min = 16 + 3 -8 - 24 +2 = - 11

Please correct me if wrong [ /hide]