Question
The greatest integer less than or equal to (√2+1)6 is
Answer: Option B
:
B
Let(√2+1)6 = k + f, where k is integral part and f
the fraction (0 ≤ f < 1).
Let(√2−1)6 = f, (0 < f < 1),
Since 0 < (√2 - 1) < 1
Now, k + f + f' =(√2+1)6 +(√2−1)6
= 2{6C0.23+6C2.22+6C4.2+6C6} = 198 ..........(i)
∴ f + f' = 198 - k = an integer
But, 0≤ f < 1 and 0 < f' < 1 ⇒ 0 < (f + f') < 2
⇒ f + f' = 1, ( ∵ f + f' is an integer)
∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.
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:
B
Let(√2+1)6 = k + f, where k is integral part and f
the fraction (0 ≤ f < 1).
Let(√2−1)6 = f, (0 < f < 1),
Since 0 < (√2 - 1) < 1
Now, k + f + f' =(√2+1)6 +(√2−1)6
= 2{6C0.23+6C2.22+6C4.2+6C6} = 198 ..........(i)
∴ f + f' = 198 - k = an integer
But, 0≤ f < 1 and 0 < f' < 1 ⇒ 0 < (f + f') < 2
⇒ f + f' = 1, ( ∵ f + f' is an integer)
∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.
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