Question

`((n+1)/n)+((n+3)/n)+((n+5)/n)+...`. The sum of the first 10 terms equals?

Answer 1

`f_10=(10n+100)/n`

Explanation:

If n is a constant then

`f_1=(n+1)/n`

`f_2=(n+1)/n+(n+3)/n=(2n+4)/n`

`f_3=(n+1)/n+(n+3)/n+(n+5)/n=(3n+9)/n`

`f_4=(n+1)/n+(n+3)/n+(n+5)/n+(n+7)/n=(4n+16)/n`

Then

`f_x=(xn+x^2)/n`

For `f_10`

`f_10=(10*n+10^2)/n=(10n+100)/n`

Answer 2

`(10n+100)/n = (10(n+10))/n`

Explanation:

The first `10` terms are

`((n+1)/n)+((n+3)/n)+((n+5)/n)+* * * +((n+19)/n)`.

Simplify to:

`(10n+(1+3+5+* * * +19))/n`

The sum of the first 10 odd numbers is `10^2 = 100` so we have

`(10n+100)/n`