Math miscellaneousHard
Question
If (4 + 
n = I + f, where n is an odd natural number. I is an integer and 0 < f < 1, then
n = I + f, where n is an odd natural number. I is an integer and 0 < f < 1, thenOptions
A.I is natural number
B.I is an even integer
C.(I + f) (1 - f) = 1
D.1 - f = (4 -)n
)nSolution
I + f (4 +)n
Let g = (4 –)n, then 0 < g < 1
I + f = nC0 + 4n + nC1 4n-1 + nC2 4n-2 15 + nC3 4n-3 ()3 + .....
g = nC0 4n - nC1 4n-1 + nC2n. 4n-2. 15 - nC3 4n-3 ()3 + ....
∴ I + f + g = 2 (nC0 4n + nC2 4n-2. 15 + .........) = even integer
∴ 0 < f + g < 2 ⇒ f + g = 1 ⇒ 1 - f = g
thus I is an odd integer
1 - f = g = (4 -)n
(I + f) (1 - f) = (I + f). g = 1
)nLet g = (4 –
)n, then 0 < g < 1I + f = nC0 + 4n + nC1 4n-1
+ nC2 4n-2 15 + nC3 4n-3 ()3 + .....g = nC0 4n - nC
+ nC2n. 4n-2. 15 - nC3 4n-3 ()3 + ....∴ I + f + g = 2 (nC0 4n + nC2 4n-2. 15 + .........) = even integer
∴ 0 < f + g < 2 ⇒ f + g = 1 ⇒ 1 - f = g
thus I is an odd integer
1 - f = g = (4 -
)n(I + f) (1 - f) = (I + f). g = 1
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