Binomial...

Algebra Level 3

If a n = ∑ r = 0 n 1 ( n r ) \displaystyle a_n=\sum_{r=0}^{n} \dfrac{1}{\binom{n}{r}} and b n = ∑ r = 0 n r ( n r ) \displaystyle b_n=\sum_{r=0}^n \dfrac{r}{\binom{n}{r}} , then find the number of ordered pairs ( p , q ) (p,q) such that c p + c q = 1 c_p+c_q=1 , where c p = a p b p c_p=\dfrac{a_p}{b_p} .

2 1 0 Infinite 4 3

Vilakshan Gupta by Vilakshan Gupta , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Oct 4, 2019

Note that b n = ∑ r = 0 n r ( n r ) = ∑ r = 0 n n − r ( n r ) = n a n − b n n ≥ 0 b_n \; = \; \sum_{r=0}^n \frac{r}{\binom{n}{r}} \; = \; \sum_{r=0}^n \frac{n-r}{\binom{n}{r}} \; = \; na_n - b_n \hspace{2cm} n \ge 0 so that n a n = 2 b n na_n = 2b_n for all n ≥ 0 n \ge 0 , and hence it follows that c n = 2 n c_n = \tfrac{2}{n} for all n ≥ 1 n \ge 1 .

Thus we want to find ordered pairs ( p , q ) (p,q) of positive integers such that 2 p − 1 + 2 q − 1 = 1 2p^{-1} + 2q^{-1} = 1 , or 2 ( p + q ) = p q 2(p+q) = pq , or ( p − 2 ) ( q − 2 ) = 4 (p-2)(q-2)=4 . Thus p − 2 p-2 can take the values 1 , 2 , 4 1,2,4 only, and there are exactly 3 \boxed{3} ordered pairs with the desired property, ( 3 , 6 ) (3,6) , ( 4 , 4 ) (4,4) and ( 6 , 3 ) (6,3) .

2 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...