Waste less time on Facebook — follow Brilliant.
×

Need derivations of some formulas - 2! Help!

  1. \(\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n \\ 1 \end{matrix} \right) +\left( \begin{matrix} n \\ 2 \end{matrix} \right) +.....+\left( \begin{matrix} n \\ n \end{matrix} \right) \quad =\quad { 2 }^{ n }\)

  2. \(\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n+1 \\ 1 \end{matrix} \right) +\left( \begin{matrix} n+2 \\ 2 \end{matrix} \right) +.....+\left( \begin{matrix} n+r \\ r \end{matrix} \right) \quad =\quad \left( \begin{matrix} n+r+1 \\ r \end{matrix} \right) \)

  3. \(\left( \begin{matrix} r \\ r \end{matrix} \right) +\left( \begin{matrix} r+1 \\ r \end{matrix} \right) +\left( \begin{matrix} r+2 \\ r \end{matrix} \right) +.....+\left( \begin{matrix} n \\ r \end{matrix} \right) \quad =\quad \left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right) \)

  4. Integrating factor - For \(\frac{dy}{dx} + Py = Q\), \(IF = {e}^{\int { P } dx}\)

Extra Credit

If you can give me some tips for such combinatorical sums.

You can add more from your own side.

Note by Kartik Sharma
2 years, 1 month ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

here (x, y) = (1 , 1)

thus it is \(( 1 + 1)^{n} = 2^{n}\)

  1. \( \binom{n + r + 1}{r} = \binom{n + r}{r} + \binom{n + r}{r - 1}\)

\( \binom{n + r }{r - 1} = \binom{n + r - 1}{r - 1} + \binom{n + r - 1}{r - 2}\)

\(\binom{n + r - 1}{r - 2} = \binom{n + r - 2}{r - 2 } + \binom{n + r - 3}{r - 3}\)

.

.

.

\(\binom{n+3}{2} = \binom{n + 2}{2} + \binom{n + 2}{1}\)

\(\binom{n+2}{1} = \binom{n + 1}{1} + \binom{n + 1}{0}\) \(= \binom{n + 1}{1} + \binom{n}{0}\)

Adding, we get,

\(\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}\)

I think now you can derive the third @Kartik Sharma

Sorry i don't know the fourth Megh Choksi · 2 years, 1 month ago

Log in to reply

@Megh Choksi Yeah, I thought of the same! Try the others! Kartik Sharma · 2 years, 1 month ago

Log in to reply

@Kartik Sharma Use fundamental formulas to derive these kinds of sums @Kartik Sharma Megh Choksi · 2 years, 1 month ago

Log in to reply

For - 1) ( Although Megh aleready post it But here one another approach for this )

Let P = Total Number's of ways of selecting n things (Say shoes ) in all possible ways

So Obviously , LHS = P

and also there is two possibility for one shoe for selection , i.e either it will be selected or not .

So all possible ways is " P " \(P\quad =\quad 2\times 2\times 2\times 2\quad .\quad .\quad .\quad .\quad .\quad .\quad 2\quad =\quad { 2 }^{ n }\quad \) Deepanshu Gupta · 2 years, 1 month ago

Log in to reply

@Deepanshu Gupta Another method for 3. ( but i think the usual is better)

Let any r of n + r + 1 objects given. let it be \( a_{1} , a_{2} ............ , a_{r}\)

Case 1 - it does not contain \(a_{1}\) , it will happen in \(\binom{n + r}{r}\) ways for the r things to choosen from the remaining n + r things.

Case 2 - it contains \(a_{1}\) but does not contain \(a_{2}\) . This will happen in \(\binom{n + r - }{ r - 1}\) ways (we have chossen \(a_{1}\) and not \(a_{2}\) , we have only n + r - 1 things to choose from and we need only r - 1)

Case 3 - it contains \( a_{1} , a_{2}\) and not \( a_{3}\) , can happen in \(\binom{n + r - 2}{r - 2}\) ways(same reasoning)

Case 4 , case 5 ...... etc

Case r - it contains \(a_{1} , a_{2} , ...... a_{r - 1}\) but not \(a_{r}\) , can happen in \(\binom{n + 1}{1}\) ways.

case r +1 - it contains all things , can happen in \(\binom{n}{0}\)ways .

Thus-

\(\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}\) Megh Choksi · 2 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...