Too much Binomial!

Algebra Level 4

( C 0 + C 1 ) ( C 1 + C 2 ) ( C 2 + C 3 ) ( C n 1 + C n ) (C_0 + C_1)(C_1 + C_2)(C_2 + C_3)\cdot \cdot \cdot (C_{n-1} + C_n)

Let C k C_k denote the binomial coefficient ( n k ) \dbinom {n}{k} , can you simplify the product above for positive integer n > 1 n > 1 ?

C 0 C 1 C 2 C 3 C n 1 ( n + 1 ) n n ! \frac{C_0C_1C_2C_3\cdot\cdot\cdot C_{n-1}(n+1)^n}{n!} C 0 C 1 C 2 C 3 C n + 1 n ! \frac{C_0C_1C_2C_3\cdot\cdot\cdot C_{n+1}}{n!} C 0 C 1 C 2 C 3 C n 1 n ! \frac{C_0C_1C_2C_3\cdot\cdot\cdot C_{n-1}}{n!} C 0 C 1 C 2 C 3 C n 2 ( n + 1 ) n n ! \frac{C_0C_1C_2C_3\cdot\cdot\cdot C_{n-2}(n+1)^n}{n!}

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.

2 solutions

Nishant Rai
Apr 20, 2015

Use the property C r n + C r 1 n = C r n + 1 \boxed{C^n_r + C^n_{r-1} = C^{n+1}_r }

Moderator note:

Can you elaborate on it?

On using the property C r n + C r 1 n = C r n + 1 \boxed{C^n_r + C^n_{r-1} = C^{n+1}_r } it gets reduced to the form C 1 n + 1 C 2 n + 1 C 3 n + 1 . . . . . C n n + 1 C^{n+1}_1 C^{n+1}_2 C^{n+1}_3 ..... C^{n+1}_n

On simplification, we get

C 1 n + 1 = ( n + 1 ) ! 1 ! n ! = ( n + 1 ) C^{n+1}_1 = \frac{(n+1)!}{1! n!} = (n+1)

C 2 n + 1 = ( n + 1 ) ! 2 ! ( n 1 ) ! = n ( n + 1 ) 2 = C 1 n ( n + 1 ) 2 C^{n+1}_2 = \frac{(n+1)!}{2!(n-1)!} = \frac{n(n+1)}{2} = \frac{C^n_1 (n+1)}{2}

C 3 n + 1 = ( n + 1 ) ! 3 ! ( n 2 ) ! = n ( n + 1 ) ( n 1 ) 3 2 = C 2 n ( n + 1 ) 3 C^{n+1}_3 = \frac{(n+1)!}{3!(n-2)!} = \frac{n(n+1)(n-1)}{3*2} = \frac{C^n_2 (n+1)}{3} and so on...

Nishant Rai - 6 years, 1 month ago

Since C n = 1 C_{n} = 1 , the first and second option are technically equal.

Shak R - 6 years, 1 month ago

Log in to reply

Thanks, I've made an addendum.

Brilliant Mathematics Staff - 6 years, 1 month ago
Rajath Naik
Apr 21, 2015

The formula that I derived is a little bit different but serves the same purpose in the end.. nCr + nC(r+1)=(n+1)C(r+1).
But not recommended for this problem..gets too hairy in the later part.lol.but still a good result to be remembered..

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...