Algebra> the positive integer is just greater than...

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

Indhumathi , 9 Years ago

Grade 11

3 Answers

Karthik Kumar

Last Activity: 9 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

So the answer is 3.

Karthik Kumar

Last Activity: 9 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

Last Activity: 7 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]