The expression (2 + sqrt{2})^4 has a value ly | vedclass

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

Question

(B) We expand $(2 + \sqrt{2})^4$ using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.$(2 + \sqrt{2})^4 = \binom{4}{0} 2^

Vedclass · Accepted Answer

$135$ and $136$

Vedclass · Answer

(B) We expand $(2 + \sqrt{2})^4$ using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.$(2 + \sqrt{2})^4 = \binom{4}{0} 2^4 + \binom{4}{1} 2^3 (\sqrt{2}) + \binom{4}{2} 2^2 (\sqrt{2})^2 + \binom{4}{3} 2^1 (\sqrt{2})^3 + \binom{4}{4} (\sqrt{2})^4$.$= 1(16) + 4(8)(\sqrt{2}) + 6(4)(2) + 4(2)(2\sqrt{2}) + 1(4)$.$= 16 + 32\sqrt{2} + 48 + 16\sqrt{2} + 4$.$= 68 + 48\sqrt{2}$.Given $\sqrt{2} \approx 1.414$,we have $48 	imes 1.414 = 67.872$.So,$68 + 67.872 = 135.872$.This value lies between $135$ and $136$.

The expression $(2 + \sqrt{2})^4$ has a value lying between

Similar Questions

If $\alpha$ and $\beta$ are the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x+\sqrt{x^{2}-1})^{6}+(x-\sqrt{x^{2}-1})^{6}$,then:

If $x$ is so small that $x^2$ and higher powers of $x$ may be neglected,then the approximate value of $\frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ is:

The coefficient of $x^{50}$ in the expansion of $(1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \ldots + x^{1000}$ is

When $x$ is so small that its square and its higher powers may be neglected,then the value of $\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}$ is approximately equal to

Using the binomial theorem,the value of $(0.999)^3$ correct to $3$ decimal places is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module