Question
The greatest integer less than or equal to (√2+1)6 is?
Answer: Option B
:
B
Let (√2+1)6=I+F, Where I is an integer and0< F<1.
f=√2−1=1√2+1⇒0<√2−1<1⇒0<f<1
Also I+F+f=(√2+1)6+(√2−1)6
= 2[6C0.23+6C2.22+6C4.2+6C6]
=2(8+60+30+1)=198
Hence, F+f=198−I is an integer. But 0<F+f<2.
∴F+f=1, and thus, I=197
Was this answer helpful ?
:
B
Let (√2+1)6=I+F, Where I is an integer and0< F<1.
f=√2−1=1√2+1⇒0<√2−1<1⇒0<f<1
Also I+F+f=(√2+1)6+(√2−1)6
= 2[6C0.23+6C2.22+6C4.2+6C6]
=2(8+60+30+1)=198
Hence, F+f=198−I is an integer. But 0<F+f<2.
∴F+f=1, and thus, I=197
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution