Let (7 + 4 
) n = Ι + f = n C 0 .7 n + n C 1 .7 n – 1 .(4 
) 1 + ....... .............(i)
where Ι & f are its integral and fractional parts respectively.
It means 0 < f < 1
Now, 0 < 7 – 4 
< 1 ⇒ 0 < (7 – 4 
) n < 1
Let (7 – 4 
) n = f ′ = n C 0 .7 n – n C 1 .7 n – 1 .(4 
) 1 + ....... .............(ii)
⇒ 0 < f ′ < 1
Adding (i) and (ii) (so that irrational terms cancelled out)
Ι + f + f ′ = (7 + 4 
) n + (7 – 4 
) n
= 2 [ n C 0 7 n + n C 2 7 n – 2 (4 
) 2 + ..........]
Ι + f + f ′ = even integer ⇒ (f + f ′ must be an integer)
0 < f + f ′ < 2 ⇒ f + f ′ = 1
with help of above analysis answer the following questions
(i)If 
= p + f, where p is an integer and f is a proper fraction, then find the value of
, n ∈ N, is
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)p + f = ( 
+ 5) n = n C 0 ( 
) n 5º + n C 1 ( 
) n – 1 5 1 + .....
f ′ = ( 
– 5) n = n C 0 ( 
) n 5º – n C 1 ( 
) n – 1 5 1 + ......
p + f + f ′ = 2 [ n C 0 ( 
) n + n C 2 ( 
) n – 2 5 2 + ......]
⇒ p + f + f ′ = even integer le (if n is even)
⇒ f + f ′ = 1 ⇒ f ′ = 1 – f
p + f – f ′ = 2 [ n C 1 ( 
) n – 1 (5) + n C 3 ( 
) n – 3 5 3 + ....]
⇒ f – f ′ = 0 ⇒ f ′ = f
(ii) 
= Ι + f
= f ′
2[ n C 0 (9) n + n C 2 (9) n–2 
+ ....] = Ι + f + f ′
∴ Ι = 2(integer ) – 1 ( f + f ′ = 1)
∴ (Ι + f) (1 – f) = 1
(iii)Let ¼ ( 
+ 1) 2n = (4 + 2 
) n = 2 n (2 + 
) n = Ι + f ..........(i)
where Ι and f are its integral & fractional parts respectively
0 < f < 1.
Now vc 0 < 
– 1 < 1
0 < ( 
– 1) 2n < 1
Let ( 
– 1) 2n = (4 – 2 
) n = 2 n (2 – 
) n = f ′ . ........(ii)
0 < f ′ < 1
adding (i) and (ii)
Ι + f + f ′ = ( 
+ 1) 2n + ( 
– 1) 2n
= 2 n [(2 + 
) n + (2 
– ) n ] = 2.2 n [ n C 0 2 n + n C 2 2 n – 2 ( 
) 2 + ........]
Ι + f + f ′ =2 n + 1 k (where k is a positive integer)
Ι + f + f ′ =2 n + 1 k
0 < f + f ′ < 2 ⇒ f + f ′ = 1
Ι + 1 = 2 n + 1 k.
Ι + 1 is the integer just above ( 
+ 1) 2n and which is divisible by 2 n + 1 .
Ι + 1 / ( 
+ 1) 2n
for ls n = 1, ( 
+ 1) 2n = ( 
+ 1) 2 ⇒ Ι + 1 = 8
so it is divisible by 8 but not by 16.
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