Question

225 can be expressed as sum of first ____ consecutive odd natural number

Answer 1

Answer:

We know that the sum of the odd numbers up to 2n−1 is n2. The sum of the odd numbers from 2m+1 through 2n−1 is then n2−m2=(n+m)(n−m) We need this to be 225, so you are looking for the number of solutions to 225=(n+m)(n−m) Each factorization of 225 gives you one except 225=15⋅15 (Why?) For example 225=45⋅5 We can then write n+m=45,n−m=5,n=25,m=20 and find 225=41+43+45+47+49. Note that there are 5 terms in the expression and the middle one is 45