If the sum of the arithmetic series $7 + 9 + 11 + 13 + 15 + . . .$ $7 + 9 + 11 + 13 + 15 + . . .$ is $25, 613, 712$ $25,613,712$ , then how many terms were added up?

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Answer 1 · 2018-03-10T17:18:18

$5058$ $5058$

Note that the sum of the first $n$ $n$ odd numbers is $n^{2}$ $n^2$

So, writing $n$ $n$ for the number of terms we are looking for:

$9+25613712 = (1+3+5)+(7+9+11+...+(2(n+3)-1))$

$color(white)(9+25613712) = (n+3)^2$

So:

$n = sqrt(9+25613712)-3 =5058$

If the sum of the arithmetic series 7 + 9 + 11 + 13 + 15 + . . .7+9+11+13+15+...7 + 9 + 11 + 13 + 15 + . . . is 25,613,71225,613,71225,613,712, then how many terms were added up?