The value of (sqrt{5} + 1)^5 - (sqrt{5} - 1)^ | vedclass

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

Question

(B) Using the binomial expansion formula $(x+a)^n - (x-a)^n = 2 \left[ {^nC_1} x^{n-1} a + {^nC_3} x^{n-3} a^3 + {^nC_5} x^{n-5} a^5 + \dots \right]$.

Vedclass · Accepted Answer

$352$

Vedclass · Answer

(B) Using the binomial expansion formula $(x+a)^n - (x-a)^n = 2 \left[ {^nC_1} x^{n-1} a + {^nC_3} x^{n-3} a^3 + {^nC_5} x^{n-5} a^5 + \dots ight]$.Here,$x = \sqrt{5}$,$a = 1$,and $n = 5$.Substituting these values:$(\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5 = 2 \left[ {^5C_1} (\sqrt{5})^4 (1)^1 + {^5C_3} (\sqrt{5})^2 (1)^3 + {^5C_5} (\sqrt{5})^0 (1)^5 ight]$$= 2 \left[ 5 	imes 25 	imes 1 + 10 	imes 5 	imes 1 + 1 	imes 1 	imes 1 ight]$$= 2 \left[ 125 + 50 + 1 ight]$$= 2 	imes 176$$= 352$.

The value of $(\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5$ is

Similar Questions

If $n$ is a positive integer,then the coefficient of $x^6$ in the expansion of $(1-2x+3x^2-4x^3+\ldots)^{-n}$ is

Let $n \geq 5$ be an integer. If $9^{n}-8n-1=64\alpha$ and $6^{n}-5n-1=25\beta$,then $\alpha-\beta$ is equal to

$(102)^4 = ?$

For $|x| < \frac{1}{\sqrt{2}}$, the coefficient of $x$ in the expansion of $\frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}$ is

Using the Binomial Theorem,evaluate $(102)^{5}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module