the sum of s

Question

The sum of S of the forst n natural numbers 1,2,3,..., n, is given by the formula S= n/2 * (n+1)

How many consecutive natural numbers starting with 1 produces the sum of 253?

 

n/2 = fraction, n over 2

                                                                                                                   

                                                                                                                        

Answer 1

S=(n/2)(n+1)

253=(n)(n+1)/2

2*253=n(n+1)

506=n^2+n

n^2+n-506=0

506=2*11*23=(2*11)*(23)=22*23 and 23-22=1

factor

(n+23)(n-22)=0

n-22=0

n=22

 

another solution...

2*253=n(n+1)

506=n(n+1) (here you want 2 consecutive numbers that multiply to 506, so take the square root of 506)

√506=22.49444..., so n=22 and n+1=23

Answer 2

The question is to solve for n, so plug in 253 for S in the formula.

 

253= (n/2)(n+1)

 

2(253)=n(n+1)

 

506 = n² + n

 

0= n² +n -506

 

0 = (n+33)(n-22)

 

n= 22;

the sum of the first 22 natural numbers is 253.