Somebody help me? I don't understand how it is solved (the factor thing is like saying x100 times)

sum of first n even nos =n(n+1)
sum of first n odd nos=n²
So
Simplify the numerator
Simplify the denominator
Sum of first 100 even nos
Sum of first 100 odd nos
So
The equation yields as
Answer:
[tex]\sf E=\dfrac{x^2 \cdot x^4 \cdot x^6 \cdot x^8 ... 100\:factors}{x \cdot x^3 \cdot x^5 \cdot x^7 ... 100\:factors}=x^{100}[/tex]
Step-by-step explanation:
Given:
[tex]\sf E=\dfrac{x^2 \cdot x^4 \cdot x^6 \cdot x^8 ...}{x \cdot x^3 \cdot x^5 \cdot x^7 ... }[/tex]
As:
[tex]\sf E=\dfrac{x^2 \cdot x^4 \cdot x^6 \cdot x^8 \cdot ... }{x \cdot x^3 \cdot x^5 \cdot x^7 \cdot ... }=\dfrac{x^2}{x^1} \cdot \dfrac{x^4}{x^3} \cdot \dfrac{x^6}{x^5} \cdot \dfrac{x^8}{x^7} \cdot ...}[/tex]
Apply the exponent rule [tex]\sf \dfrac{a^b}{a^c}=a^{b-c}[/tex]
[tex]\begin{aligned}\sf \implies \dfrac{x^2}{x^1} \cdot \dfrac{x^4}{x^3} \cdot \dfrac{x^6}{x^5} \cdot \dfrac{x^8}{x^7} \cdot... & = \sf x^{(2-1)} \cdot x^{(4-3)} \cdot x^{(6-5)} \cdot x^{(8-7)} \cdot ...\\ & = \sf x^1 \cdot x^1 \cdot x^1 \cdot x^1 \cdot ...\end{aligned}[/tex]
As there are 100 factors, then [tex]\sf x^1[/tex] is multiplied by itself 100 times ⇒ [tex]\sf x^{100}[/tex]
Therefore:
[tex]\sf E=\dfrac{x^2 \cdot x^4 \cdot x^6 \cdot x^8 ... 100\:factors}{x \cdot x^3 \cdot x^5 \cdot x^7 ... 100\:factors}=x^{100}[/tex]