Question

the integral part of root 2 + 1 whole power 6 is​

Answer 1

Answer:

Let (

2

+1)

6

=I+F where I is the integral part.

Then 0 < F < I.

Let f(

2

−1)

6

0<

2

−1<1

⇒0<(

2

−1)

6

<1⇒0<f<1

0<F<1,0<f<1⇒0<F+f<2

I+F+f(

2

+1)

6

+(

2

−1)

6

=2[

6

C

0

(

2

)

6

+

6

C

2

(

2

)

4

+

6

C

4

(

2

)

2

+

6

C

6

]

=1[8+15(4)+15(2)+1]

=2×99=198

I + F + f, I are integers

⇒ F + f is an integer.

0 < F + f < 2, F + f is an integer

⇒F+f=1

I+F+f=198⇒I+1=198

⇒I=197