Important theorem

Binomial Theorem of Class 11

An Important Theorem

If Important theorem = I + f, where I and n are positive integers n being odd.

0 ≤ f < 1, then show that (I + f) (1 − f) = Kn, where P – Q2 = K and Important theorem.

Proof:

Given Important theorem - Q < 1 ⇒ (Important theorem - Q)n < 1

Let (Important theorem - Q)n = f ′ 0 ≤ f ′ < 1

Then I + f − f′ = (Important theorem + Q)n – (√P - Q)n

RHS contain even power of Important theoremon expanding

Hence RHS and I are integers.

∴ f - f′ are integers ⇒ f - f′ = 0 as -1 < f - f′ < 1

or (I + f)f = (I + f)f ′ = (Important theorem - Q)n(Important theorem + Q)n = (P – Q2)n

Note if n is even integer

then (Important theorem + Q)n + (Important theorem - Q)n = I + f + f ′

⇒ f + f′ is also an integer

Obviously f + f′ = 1 ⇒ f′ = 1 – f

Hence (I + f)(1 – f) = (I + f) f′ = (Important theorem + Q)n (Important theorem - Q)n

= (P – Q2)n = Kn

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