Important theorem

Binomial Theorem of Class 11

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 ⇒ ( - 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 theorem on 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( + Q)n = (P – Q2)n

Note if n is even integer

then ( Important theorem + Q)ⁿ + ( - 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 ( - Q)n

= (P – Q₂)ⁿ = Kⁿ