Question

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

Answer 1

sum of first n even nos =n(n+1)

sum of first n odd nos=n²

So

Simplify the numerator

• x²x⁴x⁶..=x^(2+4+6..100times )

Simplify the denominator

• xx³x⁵..=x^{1+3+5..100times)

Sum of first 100 even nos

• 100(100-1)=100(99)=9900

Sum of first 100 odd nos

• 100²=10000.

So

The equation yields as

• E=x⁹⁹⁰⁰/x¹⁰⁰⁰⁰
• E=1/x¹⁰⁰

Answer 2

`\sf E=(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)`

Explanation:

Given:

`\sf E=(x^2 \cdot x^4 \cdot x^6 \cdot x^8 ...)/(x \cdot x^3 \cdot x^5 \cdot x^7 ... )`

As:

`\sf E=(x^2 \cdot x^4 \cdot x^6 \cdot x^8 \cdot ... )/(x \cdot x^3 \cdot x^5 \cdot x^7 \cdot ... )=(x^2)/(x^1) \cdot (x^4)/(x^3) \cdot (x^6)/(x^5) \cdot (x^8)/(x^7) \cdot ...}`

Apply the exponent rule
`\sf (a^b)/(a^c)=a^(b-c)`

`\begin{aligned}\sf \implies (x^2)/(x^1) \cdot (x^4)/(x^3) \cdot (x^6)/(x^5) \cdot (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}`

As there are 100 factors, then
`\sf x^1` is multiplied by itself 100 times ⇒
`\sf x^(100)`

Therefore:

`\sf E=(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)`