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Exercise 7.1 · Q10

Q.Using Binomial Theorem, indicate which number is larger (1.1)10000(1.1)^{10000} or 10001000.

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The key idea is to use the Binomial Theorem to expand (1.1)10000(1.1)^{10000} and show that even the first few terms already exceed 10001000. The result is that (1.1)10000(1.1)^{10000} is far larger than 10001000.

Why This Approach Works

When comparing a huge power like (1.1)10000(1.1)^{10000} with a modest number like 10001000, your first instinct might be to think: "1.1 is just a little above 1, so raising it to a high power might still be small." That intuition is wrong — and dangerously so. Exponential growth is deceptive: even a 10% increase per step, repeated 10,000 times, produces a number so large it dwarfs 1000.

The Binomial Theorem lets us write (1.1)10000(1.1)^{10000} as (1+0.1)10000(1 + 0.1)^{10000} and expand it into a sum of positive terms. We don't need the whole sum — just enough terms to prove it's already bigger than 1000.

(1+x)n=∑k=0n(nk)xk(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k

Here n=10000n = 10000 and x=0.1x = 0.1.

Step-by-Step Reasoning

  1. Write the expansion

(1.1)10000=(1+0.1)10000=(100000)+(100001)(0.1)+(100002)(0.1)2+⋯+(1000010000)(0.1)10000(1.1)^{10000} = (1 + 0.1)^{10000} = \binom{10000}{0} + \binom{10000}{1}(0.1) + \binom{10000}{2}(0.1)^2 + \cdots + \binom{10000}{10000}(0.1)^{10000}

Every term is positive, so the sum is greater than any of its parts.

  1. Look at the first two terms The first term is 11. The second term is (100001)×0.1=10000×0.1=1000\binom{10000}{1} \times 0.1 = 10000 \times 0.1 = 1000. So already:

(1.1)10000>1+1000=1001(1.1)^{10000} > 1 + 1000 = 1001

That's it — we've already beaten 1000.

  1. But wait — is that enough? The inequality is strict because there are many more positive terms after the second one. Even if we ignore them, the sum of the first two terms alone is 10011001, which is greater than 10001000. …

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