Question

1 + 4 + 7 + 10 ... what is last number that makes sum go over 1 million.

Answer 1

The sum can be expressed as


`\displaystyle\sum_(k=0)^n(3k+1)=n+1+3\sum_(k=1)^nk=n+1+\frac{3n(n+1)}2`

`=\frac32n^2+\frac52n+1`

The sum will exceed 1 million for
`n` satisfying


`\frac32n^2+\frac52n+1>1000000`

`3n^2+5n+2>2000000`

`3n^2+5n-1999998>0`

The least integer that satisfies this is
`n=816`.