*Notes to a video lecture on http://www.unizor.com*

__Algebra+ 03__

*Problem A*

Let

*be any natural number/*

**n**Define two functions of

*:*

**n**

**F(n) = 1 − 1/2 + 1/3 -…+**

+ 1/(2n-1) − 1/2n+ 1/(2n-1) − 1/2n

and

**G(n) = 1/(n+1) + 1/(n+2) +…+**

+ 1/(2n-1) + 1/2n+ 1/(2n-1) + 1/2n

Prove that for any natural

*the following equality is true:*

**n***.*

**F(n)=G(n)**For example, you can manually check the following equalities:

for

*(*

**n=1***):*

**2n=2***1 − 1/2 = 1/2*

for

*(*

**n=2***):*

**2n=4***1 − 1/2 + 1/3 − 1/4 =*

= 1/3 + 1/4

= 1/3 + 1/4

for

*(*

**n=3***):*

**2n=6***1− 1/2 + 1/3 − 1/4 +*

+ 1/5 − 1/6 =

= 1/4 + 1/5 + 1/6

+ 1/5 − 1/6 =

= 1/4 + 1/5 + 1/6

*Answer A*

Here is a straight forward proof by induction.

Check it for

*:*

**n=1**

**F(1) = 1−1/2 = 1/2**

**G(1) = 1/2**As we see,

**F(1) = G(1)**Assume,

*.*

**F(n)=G(n)**Let's switch to

**n+1**

**F(n+1) =**

= F(n) + 1/(2n+1) − 1/(2n+2)= F(n) + 1/(2n+1) − 1/(2n+2)

**G(n+1) = G(n) − 1/(n+1) +**

+ 1/(2n+1)+1/(2n+2)+ 1/(2n+1)+1/(2n+2)

When we switch from

*to*

**n***, function*

**n+1***has changed by*

**F(n)**

**f(n+1) = 1/(2n+1) − 1/(2n+2)**So,

*.*

**F(n+1)=F(n)+f(n+1)**Analogously, function

*has changed by*

**G(n)**

**g(n+1) = −1/(n+1) +**

+ 1/(2n+1) + 1/(2n+2)+ 1/(2n+1) + 1/(2n+2)

So,

*.*

**G(n+1)=G(n)+g(n+1)**If

*, an increment of*

**f(n+1)***, and*

**F(n)***, an increment of*

**g(n+1)***, are equal, assuming*

**G(n)***, we can conclude that*

**F(n)=G(n)***, which proves that*

**F(n+1)=G(n+1)***for any natural number*

**F(n)=G(n)***.*

**n**Let's check the equality

**f(n+1) = g(n+1)**Indeed, transforming an expression to a common denominator, we get

*[*

**f(n+1) = 1/***]*

**(2n+1)·(2n+2)**For

*the common denominator is*

**g(n+1)**

**(n+1)(2n+1)(2n+2)**The numerator is

**−(2n+1)(2n+2)+(n+1)(2n+2)+**

+(n+1)(2n+1) =

= −4n² − 2n − 4n − 2 +

+ 2n² + 2n + 2n + 2 +

+ 2n² + n + 2n + 1 =

= n+1+(n+1)(2n+1) =

= −4n² − 2n − 4n − 2 +

+ 2n² + 2n + 2n + 2 +

+ 2n² + n + 2n + 1 =

= n+1

Dividing this numerator of

*by its denominator obtained above and cancelling*

**g(n+1)***from both, we will get*

**n+1***[*

**g(n+1) = 1/***]*

**(2n+1)(2n+2)**This is exactly what we obtained for

*.*

**f(n+1)**END OF PROOF

*Problem B*

Two cars are approaching an intersection on two perpendicular roads.

In the beginning at time

*car #1 is at distance*

**t=0***from the intersection.*

**a**The car #2 (on a perpendicular road) at

*is at distance*

**t=0***from the intersection.*

**b**Both cars move with the same speed

*.*

**V**For traffic safety, assume that

*, so there will be no collision.*

**a≠b**At what time the distance between the cars will be minimal and what that minimal distance will be?

*Answer B*

Let

*be a distance between the cars at any moment*

**d(t)***.*

**t**Then

**d²(t) = (a−v·t)² + (b−v·t)²**Minimum of

*and*

**d(t)***occur for the same moment in time*

**d²(t)***.*

**t**So, our task is to find minimum of

*, which is a quadratic function of*

**d²(t)***.*

**t**The time when the cars are closest to each other is

**t**_{min}= (a+b)/(2V)The closest distance between the car at that time is

**d**_{min}= |a−b| /√2
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