Practice questions based on combinatorics for RMO Training and IITJEE Mathematics

Question 1:

Prove that if n is an even integer, then

\frac{1}{(1!)(n-1)!} + \frac{1}{3! (n-3)!} + \frac{1}{5! (n-5)!} + \ldots + \frac{1}{(n-1)! 1!} = \frac{2^{n-1}}{n!}

Question 2:

If {n \choose 0}, {n \choose 1}, {n \choose 2}, ….{n \choose n} are the coefficients in the expansion of (1+x)^{n}, when n is a positive integer, prove that

(a) {n \choose 0} - {n \choose 1}  + {n \choose 2} - {n \choose 3} + \ldots + (-1)^{r}{n \choose r} = (-1)^{r}\frac{(n-1)!}{r! (n-r-1)!}

(b) {n \choose 0} - 2{n \choose 1} + 3{n \choose 2} - 4{n \choose 3} + \ldots + (-1)^{n}(n+1){n \choose n}=0

(c) {n \choose 0}^{2} - {n \choose 1}^{2} + {n \choose 2}^{2} - {n \choose 3}^{2} + \ldots + (-1)^{n}{n \choose n}^{2}=0, or (-1)^{\frac{n}{2}}{n \choose {n/2}}, according as n is odd or even.

Question 3:

If s_{n} denotes the sum of the first n natural numbers, prove that

(a) (1-x)^{-3}=s_{1}+s_{2}x+s_{3}x^{2}+\ldots + s_{n}x^{n-1}+\ldots

(b) 2(s_{1}s_{2m} + s_{2}s_{2n-1} + \ldots + s_{n}s_{n+1}) = \frac{2n+4}{5! (2n-1)!}

Question 4:

If q_{n}=\frac{1.3.5.7...(2n-1)}{2.4.6.8...2n}, prove that

(a) q_{2n+1}+q_{1}q_{2n}+ q_{2}q_{2n-1} + \ldots + q_{n-1}q_{n+2} + q_{n}q_{n+1}= \frac{1}{2}

(b) 2(q_{2n}-q_{1}q_{2m-1}+q_{2}q_{2m-2}+\ldots + (-1)^{m}q_{m-1}q_{m+1}) = q_{n} + (-1)^{n-1}{q_{n}}^{2}.

Question 5:

Find the sum of the products, two at a time, of the coefficients in the expansion of (1+x)^{n}, where n is a positive integer.

Question 6:

If (7+4\sqrt{3})^{n} = p + \beta, where n and p are positive integers, and \beta, a proper fraction, show that (1-\beta)(p+\beta)=1.

Question 7:

If {n \choose 0}, {n \choose 1}, {n \choose 2}, …,, {n \choose n} are the coefficients in the expansion of (1+x)^{n}, where n is a positive integer, show that

{n \choose 1} - \frac{{n \choose 2}}{2} + \frac{{n \choose 3}}{3} - \ldots + \frac{(-1)^{n-1}{n \choose n}}{n} = 1 + \frac{1}{2} + \frac {1}{3} + \frac{1}{4} + \ldots + \frac{1}{n}.

That’s all for today, folks!

Nalin Pithwa.

 

 

 

 

 

 

 

 

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