# 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.

Note by Kartik Sharma
3 years, 7 months ago

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

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

1. $$\binom{n + r + 1}{r} = \binom{n + r}{r} + \binom{n + r}{r - 1}$$

$$\binom{n + r }{r - 1} = \binom{n + r - 1}{r - 1} + \binom{n + r - 1}{r - 2}$$

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

.

.

.

$$\binom{n+3}{2} = \binom{n + 2}{2} + \binom{n + 2}{1}$$

$$\binom{n+2}{1} = \binom{n + 1}{1} + \binom{n + 1}{0}$$ $$= \binom{n + 1}{1} + \binom{n}{0}$$

$$\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}$$

I think now you can derive the third @Kartik Sharma

Sorry i don't know the fourth

- 3 years, 7 months ago

Yeah, I thought of the same! Try the others!

- 3 years, 7 months ago

Use fundamental formulas to derive these kinds of sums @Kartik Sharma

- 3 years, 7 months ago

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 " $$P\quad =\quad 2\times 2\times 2\times 2\quad .\quad .\quad .\quad .\quad .\quad .\quad 2\quad =\quad { 2 }^{ n }\quad$$

- 3 years, 7 months ago

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-

$$\binom{n}{0} + \binom{n + 1}{1} + ............. + \binom{n + r}{r} = \binom{n + r + 1}{r}$$

- 3 years, 7 months ago