Question

Can you help me solving this in detail ?​

Answer 1

Answer:

Solution :-

$\longrightarrow\bigg(1 + \dfrac{1}{x}\bigg) \bigg(1 + \dfrac{1}{x + 1}\bigg) \bigg(1 + \dfrac{1}{x + 2}\bigg) \bigg(1 + \dfrac{1}{x + 3}\bigg)$

$\Rightarrow\bigg(\dfrac{{x}^{2} + 1}{x}\bigg) \bigg(\dfrac{x + 1 + 1}{x + 1}\bigg) \bigg(\dfrac{x + 2 + 1}{x + 2}\bigg) \bigg(\dfrac{x + 3 + 1}{x + 3}\bigg)$

$\Rightarrow\dfrac{{x}^{2} + x}{x} \times \dfrac{x + 2}{x + 1} \times \dfrac{x + 3}{x + 2} \times \dfrac{x + 4}{x + 3}$

$\Rightarrow\dfrac{{x}^{2} + x}{x} \times \dfrac{x + 4}{x + 1}$

$\Rightarrow\dfrac{x^3 + 4x^2 + x^2 + 4x}{x^2 + 1}$

$\Rightarrow\dfrac{x^3 + 5x^2 + 4x}{x^2 + 1}$

$\Rightarrow\dfrac{x(x^2 + 5x + 4)}{x^2 + 1}$

$\Rightarrow\dfrac{x\{x^2 + (4 + 1)x + 4\}}{x^2 + 1}$

$\Rightarrow\dfrac{x\{x^2 + 4x + x + 4\}}{x^2 + 1}$

$\Rightarrow\dfrac{x\{x(x + 4) + 1(x + 4)\}}{x^2 + 1}$

$\Rightarrow\dfrac{x(x + 4) (x + 1)}{x^2 + 1}$

$\therefore The\: answer\: of\: this\: question\: is\:\dfrac{x(x + 4) ( x + 1)}{x^2 + 1}$