Question

1/n+1- 1/n+2= 18/9n+3​

Answer 1

Step-by-step explanation:

1/(n+1) -1/(n+2) = 18/9n+3 =>

(n+2)-(n+1)/(n+1)(n+2) = 6/3n+1 =>

1/(n+1)(n+2) = 6/3n+1 => 6(n+1)(n+2)-

(3n+1) = 0 =>

$6( {n}^{2} + 3n + 2) - (3n + 1) = 0 = >$

$6 {n}^{2} + 18n + 12 - 3n - 1 = 0 = >$

$6 {n}^{2} + 15n + 11 = 0 = >$

$n = \frac{ - 15 + \sqrt{225 - 4 \times 66} }{12} \: and$

$n = \frac{ - 15 - \sqrt{225 - 4 \times 66} }{12}$

=> n = (-15+√39i)/12 and (-15-√39i)/12. So to has no real roots but only imaginary roots.

Hope it helps you.