The city of Leighton is installing street lights in a neighborhood that currently lacks street lights.

Question

 For the next fifty two weeks, they will install five new lights each week in this neighborhood, and these will be in operation starting the week after installation. Each light costs $1 per week to operate.

 

How many lights will be in operation during week n? check for n=1,n=2 and n=52.

What is the cost of operation for the lights for week n?

What is the total cost of operation incurred over these 52 weeks?

Answer 1

 

F(n) = {   0 if n=0,

                5 if n = 1,

         

               F(n-1)+5 otherwise

 

         }

 

So for n>1

 

F(1) = 5

F(2) = F(1) + 5 = 5+5 = 2*5

F(3) = F(2) + 5 = 2*5 + 5 = 3*5

...

For positive integer k,

F(k) = 5k

 

F(k+1) = F(k) + 5 = 5K + 5 = 5(k+1) <--- this completes the proof by induction

 

So F(n) = 5n for any integer n

 

Since it is $1 per week, the cost shall be $5n

 

Finally, we want the sum of of 5(1) + 5(2) + 5(3) + .... + 5(50) + 5(51) + 5(52)

 

 

     Factoring out the 5,   5 ( 1 + 2 + 3 + .... + 50 + 51 + 52) 

 

   The sum of the first k integers is k(k+1)/2, as proven below.

 So the total cost is 5 ( 52 * 53/2) = 5 * 26 * 53 = 6890

 

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Theorem: Sum of the first N positive integers 1 + 2+3 + ... + N = N(N+1)/2

 

Proof by induction:

 

   Sum of the first few whole numbers are  N=1:  1 = 1*2/2 

                                                              N=2:  1 + 2 = 3 = 2*3/2

                                                              N=3:  1 + 2 + 3 = 6 = 3*4/2

                                                              N=4 :  1 +2 +3 +4 = 10 = 4*5/2

 

Given 1 + 2 + 3 + .... + k = k(k+1)/2 for positive integer k>4;

 

Then 1 +2 + 3 + ... + k + (k + 1) = k(k+1)/2 + k+1

 

                                                   = k(k+1)/2 + 2(k+1)/2 <--- common denominator

 

                                                   = [k(k+1) + 2(k+1)]/2 <--- adds the numerators

 

                                                     = [k^2 + k + 2k + 2 ]/2 <--- distributive 

 

                                                     = (k^2 + 3x + 2)/2  <--- combines like terms

 

                                                     = ( k+1      )(k+2       )/2

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