Binomial TheoremHard
Question
If (9 + 
)n = I + f, where I, n are integers and 0 < f < 1, then :
)n = I + f, where I, n are integers and 0 < f < 1, then :Options
A.I is an odd integer
B.I is an even integer
C.(I + f) (1 - f) = 1
D.1 - f = (9 - 
)n
)nSolution
(9 - )n = I + f
(9 -)n = f′
2[nC0 (9)n + nC2 (9)n-2 (
)2 + .... ] = I + f + f′
∴ I = 2(integer) - 1 (∵ f + f′ = 1)
∴ (I + f) (1 - f) = 1
)n = I + f(9 -
)n = f′2[nC0 (9)n + nC2 (9)n-2 (
)2 + .... ] = I + f + f′ ∴ I = 2(integer) - 1 (∵ f + f′ = 1)
∴ (I + f) (1 - f) = 1
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