Question regarding binomial theorem.

[sankalpmittal]

Homework Statement

(√2 + 1)⁶ = I + f

Where I is the sum of integer part of the expansion of (√2 + 1)⁶ and f is sum of the fraction part in (√2 + 1)⁶.

Homework Equations

(x+1)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 + ⁿC₂ x^n-2 + ... + ⁿC_n

ⁿC_n = ⁿC₀ = 1

The Attempt at a Solution

I expanded (√2 + 1)⁶ , then simplified and then got the expression 44+99√2/2.
Then I got I=44 which was not even the correct answer. The correct was 197. This question is a competitive level question.

Thanks in advance...

 

[Infinitum]

Hi sankalp

The answer you get is incomplete. The second term also has an integral and fractional part itself. You need to add them to get your answer

Another way to see it is that f is defined to be between 0 and 1, so is your fractional part between zero and one?PS : Your expansion itself seems incorrect to me. Recheck it.

 

[sankalpmittal]

Hi sankalp

The answer you get is incomplete. The second term also has an integral and fractional part itself. You need to add them to get your answer

Another way to see it is that f is defined to be between 0 and 1, so is your fractional part between zero and one?


PS : Your expansion itself seems incorrect to me. Recheck it.

Hii Infinitum!

One way to do is to seriously find (√2 + 1)⁶. But that will be a noob way.

Ok , so on expanding , I get :

(√2 + 1)⁶ = ⁶C₀ 8 + ⁶C₁ 4√2 + ⁶C₂ 4 + ⁶C₃2√2 + ⁶C₄ 2 + ⁶C₅ √2 + ⁶C₆

Now on solving , I get (√2 + 1)⁶ = 99+ 70√2
I + f = 99+ 70√2

Now what else can I do ? Any hint ?

 

[Infinitum]

Hii Infinitum!

One way to do is to seriously find (√2 + 1)⁶. But that will be a noob way.

Ok , so on expanding , I get :

(√2 + 1)⁶ = ⁶C₀ 8 + ⁶C₁ 4√2 + ⁶C₂ 4 + ⁶C₃2√2 + ⁶C₄ 2 + ⁶C₅ √2 + ⁶C₆

Now on solving , I get (√2 + 1)⁶ = 99+ 70√2
I + f = 99+ 70√2

Now what else can I do ? Any hint ?

Yep. That sounds correct. Now you can use the approximate value of √2 to multiply, and hence get the integral part of the expansion.

Hint : You only need to use 1.41 as your approximation, as any more digits will not change effect the integral part

 

[sankalpmittal]

Yep. That sounds correct. Now you can use the approximate value of √2 to multiply, and hence get the integral part of the expansion.

Did not recognize that this was so simple...

99+ 70√2

99+ 70(1.41)
I = 99+98 = 197 !

Awesome!

Edit : All right , thanks for the efforts...

 

[Infinitum]

The method your professor uses can be applied in general to all such problems, so it is good in its own right. The one I suggest requires that you know the value of √2, which is frequently used and known. What about when it is √327? You would have to first find the square root, then approximate.