Do you know the power of 1000?

By using the Binomial Approximation , we know:

$(1 + b)^n \approx 1+nb$ , then:

$(1 + 0.0001)^{1000} \approx 1+0.0001 \times 1000 = 1.1 < 2$

We know that $(1+10^{-4})^{10^4} \approx e$ , so $(1+0.0001)^{1000} \approx e^{0.1}$ . Since $e$ is slightly less than 3, $\sqrt{e}$ must be less than 2 already; $e^{0.1}$ is even smaller, and obviously greater than 1, so the answer is 2.

Suppose that you don't have a calculator to hand. We must create a range for the value given. Firstly we will expand the bracket, using the Binomial Theorem (note that 0.0001 = $\frac{1}{10^4}$ ): $(1+0.0001)^{1000}$ = 1 + ${1000\choose1}$ . $\frac{1}{10^4}$ + ${1000\choose2}$ . $\frac{1}{10^8}$ +...+ $\frac{1}{10^4000}$ . Now this is ${more}$ ${than}$ 1 + $\frac{1}{10^4}$ + $\frac{1}{10^8}$ +... = $\displaystyle \sum_{i=0}^\infty$ $\frac{1}{10^4}$ = $\frac{10000}{9999}$ . It is also ${less}$ ${than}$ : $\displaystyle \sum_{i=0}^\infty$ $\frac{1}{10^i}$ = $\frac{10}{9}$ . In other words: $\frac{10000}{9999}$ < $(1+0.0001)^{1000}$ < $\frac{10}{9}$ , so the next integer is $\boxed{2}$ .

1 to the power nearly anything is 1. So is any number closing in around 1. 1+0.0001 gives you 1.0001^1000. Which would be around 1.00000000000000000000.........00011. The integer after this number is 2.

f(x+∆x)=f(x)+f'(x)*∆x when ∆x->0. Taking f(x)=x^1000 and ∆x=0.0001 we get f(1.0001)=1.1 .Thus the answer is 2.

System.out.println ( Math.ceil (Math.pow (1.0001,1000) ) ) ;

I used this computer program in java, just one small line does the work, no need for binomial approximation or range calculation needed.

Using the basics of the binomial theorem, we can see that the expression (1 + 0.0001)^1000 will be equal to:

(1^1000 * 0.0001^0) + (1^999 * 0.0001^1) + (1^998 * 0.0001^2) ... + (1^0 * 0.0001^1000)

The first term of each expression is simply 1, so we can "factor out" the 1 to make the new expression:

1(0.0001^0 + 0.0001^1 + 0.0001^2 + ... + 0.0001^1000)

We can easily see that besides the first number 0.0001^0, which is 1, the rest of the numbers will be extremely small approaching closer and closer to 0. Thus, the rest of the numbers are irrelevent except in showing us that they sum to be a number slightly greater than 0 but definitely less than 1.

1(1 + very small number) = 1.0001 with the 0001 repeating. The smallest positive integer greater than that expression is simply 2.

1+1000*0.0001=1.1 hence 2

By binomial approximation- (1+x)^n=1+nx, when |x|<1. so, (1+0.0001)^1000 =1+(1000)(0.0001) =1.1 Therefore the required solution is 2.

An easy way to think of this is to compare it to what happens if you multiply a number, say 1, by 1.01. It becomes 1.01 - or increases by 10%. In this case, we are multiplying 1 by 1.0001 (a 0.1% increase) 1000 times (if you count adding 0.0001 to 1 as being equivalent, which it is)... if you apply a 0.1% increase 1000 times you end up with a 100% increase. The number which is 100% greater than 1 is 2.