If (3+sqrt{2})^6-(3-sqrt{2})^6=a+b sqrt{2},th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \left[ \binom{n}{1} x^{n-1} y + \binom{n}{3} x^{n-3} y^3 + \binom{n}{5} x^{n-5} y^5 \right]$.H

Vedclass · Accepted Answer

$5220$

Vedclass · Answer

(A) Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \left[ \binom{n}{1} x^{n-1} y + \binom{n}{3} x^{n-3} y^3 + \binom{n}{5} x^{n-5} y^5 \right]$.Here $x=3, y=\sqrt{2}, n=6$.$(3+\sqrt{2})^6 - (3-\sqrt{2})^6 = 2 \left[ \binom{6}{1} (3)^5 (\sqrt{2}) + \binom{6}{3} (3)^3 (\sqrt{2})^3 + \binom{6}{5} (3)^1 (\sqrt{2})^5 \right]$.$= 2 \left[ 6 \cdot 243 \cdot \sqrt{2} + 20 \cdot 27 \cdot 2\sqrt{2} + 6 \cdot 3 \cdot 4\sqrt{2} \right]$.$= 2 \left[ 1458\sqrt{2} + 1080\sqrt{2} + 72\sqrt{2} \right]$.$= 2 \left[ 2610\sqrt{2} \right] = 5220\sqrt{2}$.Comparing with $a+b\sqrt{2}$,we get $a=0$ and $b=5220$.Therefore,$a+b = 0 + 5220 = 5220$.

If $(3+\sqrt{2})^6-(3-\sqrt{2})^6=a+b \sqrt{2}$,then $a+b=$

Similar Questions

The value of the expression $\frac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}$ at $x = 2$ is:

If the ratio of the coefficients of the third and fourth terms in the expansion of $(x - \frac{1}{2x})^n$ is $1 : 2$,then the value of $n$ is:

Consider the data on $x$ taking the values $2^0, 2^1, 2^2, \ldots, 2^n$ with frequencies ${}^nC_0, {}^nC_1, {}^nC_2, \ldots, {}^nC_n$ respectively. If the mean of this data is $\frac{728}{2^n}$,then $n$ is equal to

In the expansion of $(1 + x)^5$,the sum of the coefficients of the terms is

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module