Question

Find n such that: 2/3+2/15+2/35+...+ 2/ nx(n+2)=322/323

Answer 1

2/3+2/15+2/35+...+2/n x (n+2)=322/323

=2/1.3+2.3.5+2/5.7+...+2/n x(n+2)=322/323

=                                         n+2 =323

                                             n=323-2

                                              n=321    

Answer 2

2/3+2/15+2/35+...+2/n x (n+2) = 322/323

2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323

1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323

1-1/n+2 = 322/323

1/n+2 = 1-322/323

1/n+2 = 1/323

=> n+2 = 323

n = 323 - 2 = 321