Question

using. binomial tjeorem (1.1)3
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Answer 1

(1.1)^(10000) = (1 + 0.1)^(10000)

use formula,

(x + y)^n = nC0.x^n + nC1.x^(n-1)y + nC2.x^(n-2)y^2 + …..+nCn.y^n

(1 + 0.1)^(10000) = 10000C0.(1)^(10000) + 10000C1.(1)^9999(0.1) + …. other terms

= 1 × 1 + 10000 × 0.1 + … other terms .

= (1 + 1000 + ….other terms ) > 1000

hence, (1.1)^(10000) > 1000