If the odd natural numbers are arranged like a pyramid , then what r the extremities of the row containing the number 2009. 1

Question

_{they r arranged like this:-.          1}

                                      3.         5

                                7.       9.        11....

 

Answer 1

For this question, it is clear that the number of items in each row increases by 1, so the number at the end of each row is given by   2*sum - 1, where sum is the sum of the first n  integers.  Since sum = n(n+1)/2, the number at the end of row  n is given by 2*(n*(n+1)/2) - 1 = n(n+1) -1

 

So let's find the end row number that is nearest but smaller than 2009

 

2009 >=  n(n+1) -1;   n2+n-2010 >= 0   

 

Solving for n and keeping only the positive root gives   -1/2 + sqrt(1 + 4*1*2010)/2  using the quadratic formula.

gives n <= 44.3

 

Looking at row 44 gives 44(45) -1 = 1979 as the final number in the row.  The next row (n=45) then starts with 1981, the next odd number after 1979.  That row has 45 odd numbers including 2009.  The number at the end of row 45 is 45(46) - 1 = 2069.

 

So the extremities are 1981 and 2069.