Question

the sum of six consecutive positive integers beginning at n is a perfect cube. the smallest such n is 2. Find the sum of the next two smalles such n's.

Answer 1

if we sum the six consecutive positive integers starting from n, we would get their sum S as:\[S=6n+15=3(2n+5)\]you are told that this is a perfect cube, so we know:\[S=m^3\]for some positive integer m. We therefore have:\[m^3=3(2n+5)\]therefore:\[2n=\frac{m^3}{3}-5\]From this, we can tell that m must be a multiple of 3, so let:\[m=3k\]to get:\[2n=\frac{27k^3}{3}-5=9k^3-5\]therefore:\[n=\frac{9k^3-5}{2}\]k=1 gives us n=2. So now just try increasing values of k to see which ones will give an integer n.

Answer 2

the geometric sum gives us n-1/n-2 , right?

Answer 3

It's not a geometric series, it's an arithmetic series.