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# the positive integer is just greater than (1+0.0001)10000 is ​ a) 4 b) 5 c) 2 d) 3

Karthik Kumar
100 Points
6 years ago
It is well known that lim n—> inf (1 + 1/n)^n evaluates to e.
So if we put n = 10000
(1 + 1/10000)^10000
And by binomial theorem,
(1 + 1/10000)^10000  = 1 + 10000C1 1/1000 + … > 2
Karthik Kumar
100 Points
6 years ago
It is well known that lim n—> inf (1 + 1/n)^n evaluates to e.
So if we put n = 10000
(1 + 1/10000)^10000
And by binomial theorem,

Using the binomial theorem:
( 1 + 1/10000 )^10000    =  10000C0  + 10000C1*(1/10000)  +  10000C2 * (1/10000)^2  + 10000C3*(1/10000)^3 + ……….. + 10000C10000 * (1/10000)*10000
=  1   +    1    +   10000C2*(1/10000)^2  +  10000C3*(1/10000^3) + …….
Now, 10000C2  = 10000*9999/2 , which is obviously less than 10000*10000/2 = (1/2)*1000^2
Similarly , 10000C3  = 10000*9999*9998 / 6   , which is less than  10000*10000*10000 / 6   = (1/6)*10000^3
and so on….
so  (1 + .0001)^10000  =  1   +   1   +   10000C2*(1/10000^2)   +  10000C3*(1/10000^3 ) + ……. + 10000C10000*(1/10000^10000)
So, (1 + .0001)^10000
==> (1 + .0001)^10000
Now, we know that 2 + 1/2 + 1/4 + 1/8 + ….    upto infinity =3
==>  (1 + .0001)^10000
2
=> [x] = 2
Akram Khan
35 Points
4 years ago
It is well known that lim n—> inf (1 + 1/n)^n evaluates to e.So if we put n = 10000(1 + 1/10000)^10000And by binomial theorem, then we have aur expansionUsing the binomial theorem:( 1 + 1/10000 )^10000 = 10000C0 + 10000C1*(1/10000) + 10000C2 * (1/10000)^2 + 10000C3*(1/10000)^3 + ……….. + 10000C10000 * (1/10000)*10000= 1 + 1 + 10000C2*(1/10000)^2 + 10000C3*(1/10000^3) + ……. According we get New=[3]